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Page No 328:

Question 1:

Answer:

The reactance of a capacitor is given by,
Xc=1ωC
For a DC source, ω = 0
Xc=10×C=
So, for a constant DC source, reactance of a capacitor is infinite.

Page No 328:

Question 2:

The reactance of a capacitor is given by,
Xc=1ωC
For a DC source, ω = 0
Xc=10×C=
So, for a constant DC source, reactance of a capacitor is infinite.

Answer:

Voltage, V = V0cos ωt
Current, i = i0 sin ωt or i = i0 cos (ω-π2)

Power dissipated in an AC circuit is given by,
P=IrmsVrmscosϕ,
where Irms = rms value of current
Vrms= rms value of voltage
Ï• = phase difference between current and voltage
Here, Ï• = π/2
cosϕ=cosπ2=0P=IrmsVrms×0=0

Page No 328:

Question 3:

Voltage, V = V0cos ωt
Current, i = i0 sin ωt or i = i0 cos (ω-π2)

Power dissipated in an AC circuit is given by,
P=IrmsVrmscosϕ,
where Irms = rms value of current
Vrms= rms value of voltage
Ï• = phase difference between current and voltage
Here, Ï• = π/2
cosϕ=cosπ2=0P=IrmsVrms×0=0

Answer:

The rms value of current is given by,
irms=i02
Since peak value of current i0 is same for both currents, their rms values will be same.

Page No 328:

Question 4:

The rms value of current is given by,
irms=i02
Since peak value of current i0 is same for both currents, their rms values will be same.

Answer:

Let a LCR circuit is connected across an AC supply with the emf E = E0 sin ωt.
Let the inductance in the circuit be L
Let the net impedence of the circuit be Z=R2+(XL-XC)2
Where,
R =  resistance in the circuit
XL= reactance due to inductor
XC= reactance due to capacitor
The magnitude of the voltage across the inductor is given by
V=Ldidt
The current in the circuit can be written as I=I0sin(ωt+ϕ)
Where, Ï• is the phase difference between the current and the supply voltage
Thus, the voltage across the inductor can be written as
V=LI0cos(ωt+ϕ)
Thus peak value of the voltage across the inductor is given by
V=LI0V=E0Z×L
Therefore, the peak voltage across the inductor is given by V=E0Z×L
At resonance Z = R,
V=E0R×L,
If LR>1
V > E0
Therefore if magnitude of LR>1 at resonance the value of the voltage across the inductor will bw greater than the peak value of the supply voltage.

Page No 328:

Question 5:

Let a LCR circuit is connected across an AC supply with the emf E = E0 sin ωt.
Let the inductance in the circuit be L
Let the net impedence of the circuit be Z=R2+(XL-XC)2
Where,
R =  resistance in the circuit
XL= reactance due to inductor
XC= reactance due to capacitor
The magnitude of the voltage across the inductor is given by
V=Ldidt
The current in the circuit can be written as I=I0sin(ωt+ϕ)
Where, Ï• is the phase difference between the current and the supply voltage
Thus, the voltage across the inductor can be written as
V=LI0cos(ωt+ϕ)
Thus peak value of the voltage across the inductor is given by
V=LI0V=E0Z×L
Therefore, the peak voltage across the inductor is given by V=E0Z×L
At resonance Z = R,
V=E0R×L,
If LR>1
V > E0
Therefore if magnitude of LR>1 at resonance the value of the voltage across the inductor will bw greater than the peak value of the supply voltage.

Answer:

Ohm's Law is valid for resistive circuits only. It is not valid for capacitive or inductive circuits, or a combination of both. 

Page No 328:

Question 6:

Ohm's Law is valid for resistive circuits only. It is not valid for capacitive or inductive circuits, or a combination of both. 

Answer:

The reactance of a capacitor is given by,
Xc=1ωC
Also,
C=Kε0Ad,
where C = capacitance
K = dielectric constant
A = area of plates
d = distance between the plates.
K>1
The capacitance C of the capacitor will increase on inserting the dielectric slab and, consequently, the reactance Xc will decrease.
Rms current, irms=ε02XC

Therefore, rms current will decrease.

Page No 328:

Question 7:

The reactance of a capacitor is given by,
Xc=1ωC
Also,
C=Kε0Ad,
where C = capacitance
K = dielectric constant
A = area of plates
d = distance between the plates.
K>1
The capacitance C of the capacitor will increase on inserting the dielectric slab and, consequently, the reactance Xc will decrease.
Rms current, irms=ε02XC

Therefore, rms current will decrease.

Answer:

The condition for resonance is:
1ωC=ωL
The peak current through the circuit is given by,
i0=V0R2+1ωC-ωL2
From the condition of resonance, we get:
i0=V0R
The current will flow through the all circuit elements. But since the reactance of the capacitor and inductor are equal, the potential difference across them will be equal and opposite and will cancel each other.

Page No 328:

Question 8:

The condition for resonance is:
1ωC=ωL
The peak current through the circuit is given by,
i0=V0R2+1ωC-ωL2
From the condition of resonance, we get:
i0=V0R
The current will flow through the all circuit elements. But since the reactance of the capacitor and inductor are equal, the potential difference across them will be equal and opposite and will cancel each other.

Answer:

No. When an AC source is connected to a capacitor, there is a steady in the circuit to transfer change to the plates of the capacitor. This produces a potential difference between the plates. The capacitance is alternatively charged and discharged as the current reverses after each half cycle.



Page No 329:

Question 9:

No. When an AC source is connected to a capacitor, there is a steady in the circuit to transfer change to the plates of the capacitor. This produces a potential difference between the plates. The capacitance is alternatively charged and discharged as the current reverses after each half cycle.

Answer:

The thermal energy produced for an AC circuit in one time period is given by,
H=Irms2×R×2πω
For current, i1 = i0 sin ωt,
Irms=i02
H=i02R2×2πω=πi02Rω

For current, i2 = −i0 sin ωt,
Irms=i02
Hence, the same thermal energy will be produced due to this current.

Since, the direction of i1 and i2 are opposite and their magnitude is same, the net current through the resistor will become zero when both are passed together. Yes, the principle of superposition is obeyed in this case.

Page No 329:

Question 10:

The thermal energy produced for an AC circuit in one time period is given by,
H=Irms2×R×2πω
For current, i1 = i0 sin ωt,
Irms=i02
H=i02R2×2πω=πi02Rω

For current, i2 = −i0 sin ωt,
Irms=i02
Hence, the same thermal energy will be produced due to this current.

Since, the direction of i1 and i2 are opposite and their magnitude is same, the net current through the resistor will become zero when both are passed together. Yes, the principle of superposition is obeyed in this case.

Answer:

When a transformer steps up the voltage, the voltage increases but current decreases. Neglecting any loss of energy, the power remains constant and, hence, energy is not produced. It remains constant.

Page No 329:

Question 11:

When a transformer steps up the voltage, the voltage increases but current decreases. Neglecting any loss of energy, the power remains constant and, hence, energy is not produced. It remains constant.

Answer:

A transformer is ideally an inductive coil. For an inductor connected across a DC voltage,

V-Ldidt=0V=Ldidtdi=VLdti=VtL

For a DC source, the current across the inductor will increase with time and can reach a very large value, which can burn the transformer.

Page No 329:

Question 12:

A transformer is ideally an inductive coil. For an inductor connected across a DC voltage,

V-Ldidt=0V=Ldidtdi=VLdti=VtL

For a DC source, the current across the inductor will increase with time and can reach a very large value, which can burn the transformer.

Answer:

Let us consider an AC series LCR circuit of angular frequency ω. The impedance of the circuit is given by,
Z=R2+ωL-1ωC2
The phase difference between V and I is given by,
tanϕ=ωL-1ωCZ
From the above formula, we can clearly see that
ϕ-π2,π2
So, we cannot have a phase difference of 180° or 120°.

Page No 329:

Question 13:

Let us consider an AC series LCR circuit of angular frequency ω. The impedance of the circuit is given by,
Z=R2+ωL-1ωC2
The phase difference between V and I is given by,
tanϕ=ωL-1ωCZ
From the above formula, we can clearly see that
ϕ-π2,π2
So, we cannot have a phase difference of 180° or 120°.

Answer:

When a capacitor is included in a series circuit, the impedance of the circuit,
Z=R2+XC2
The power absorbed by a resistor is given by,
P=Irms2R
Since impedance increases due to introduction of a capacitor, the rms value of current Irms will decrease and, hence, the power absorbed by the resistor will decrease.

When a small inductance is introduced in the circuit, the impedance of the circuit,
Z=R2+XC-XL2
Since the impedance now decreases a little, the rms value of current will increase and, hence, the power absorbed by the resistor will increase.

Page No 329:

Question 14:

When a capacitor is included in a series circuit, the impedance of the circuit,
Z=R2+XC2
The power absorbed by a resistor is given by,
P=Irms2R
Since impedance increases due to introduction of a capacitor, the rms value of current Irms will decrease and, hence, the power absorbed by the resistor will decrease.

When a small inductance is introduced in the circuit, the impedance of the circuit,
Z=R2+XC-XL2
Since the impedance now decreases a little, the rms value of current will increase and, hence, the power absorbed by the resistor will increase.

Answer:

A hot-wire ammeter measures the rms value of current for an alternating current. So, it can be used to measure the direct current of constant value because that constant value will be equal to the rms value of current. As, the rms value of the current is same as the direct current thus we need not change the graduations.

Page No 329:

Question 1:

A hot-wire ammeter measures the rms value of current for an alternating current. So, it can be used to measure the direct current of constant value because that constant value will be equal to the rms value of current. As, the rms value of the current is same as the direct current thus we need not change the graduations.

Answer:

(a) DC

For a DC source, ω = 0. So, the reactance of the capacitance is given by,
XC=1ωC=10=

Page No 329:

Question 2:

(a) DC

For a DC source, ω = 0. So, the reactance of the capacitance is given by,
XC=1ωC=10=

Answer:

(c) i1 < i2

The charge on the capacitor during steady state is given by,
Q=Cε=ε0Ccos100πs-1t+cos500πs-1t

The steady state current is, thus, given by,
i=dQdt=ε0C×100πsin100πs-1t+ε0C×500πsin500πs-1ti=100Cπε0 cos100πs-1t+ϕ1+500Cπε0 cos500πs-1+ϕ2i1=100Cπε0 & i2=500Cπε0i2>i1

Page No 329:

Question 3:

(c) i1 < i2

The charge on the capacitor during steady state is given by,
Q=Cε=ε0Ccos100πs-1t+cos500πs-1t

The steady state current is, thus, given by,
i=dQdt=ε0C×100πsin100πs-1t+ε0C×500πsin500πs-1ti=100Cπε0 cos100πs-1t+ϕ1+500Cπε0 cos500πs-1+ϕ2i1=100Cπε0 & i2=500Cπε0i2>i1

Answer:

(c) about 310 V

Given:
Vrms = 220 V
The peak value of voltage is given by,
Vp=2×Vrms=1.414×220=311 V

Page No 329:

Question 4:

(c) about 310 V

Given:
Vrms = 220 V
The peak value of voltage is given by,
Vp=2×Vrms=1.414×220=311 V

Answer:

(b) may be zero

Let the AC voltage be given by,
V=V0sinωt
Here, ω = 2πf = 314 rad/s
The average voltage over the given time,
Vavg=00.01Vdt00.01dt=-V0cosωtω00.01         =V0ω×0.011-cosω0.01         =V0314×0.011-cos314×0.01         =V03.141-cosπ         =2V0π=140.127 V

Also, when V=V0cosωt,
Vavg=00.01Vdt00.01dt=V0sinωtω00.01         =V0ω×0.01sinω0.01-0         =V0314×0.01sin314×0.01         =V03.14sinπ         =0
   
From the above results, we can say that the average voltage can be zero. But it is not necessary that it should be zero or never zero.

Page No 329:

Question 5:

(b) may be zero

Let the AC voltage be given by,
V=V0sinωt
Here, ω = 2πf = 314 rad/s
The average voltage over the given time,
Vavg=00.01Vdt00.01dt=-V0cosωtω00.01         =V0ω×0.011-cosω0.01         =V0314×0.011-cos314×0.01         =V03.141-cosπ         =2V0π=140.127 V

Also, when V=V0cosωt,
Vavg=00.01Vdt00.01dt=V0sinωtω00.01         =V0ω×0.01sinω0.01-0         =V0314×0.01sin314×0.01         =V03.14sinπ         =0
   
From the above results, we can say that the average voltage can be zero. But it is not necessary that it should be zero or never zero.

Answer:

(b) 50 Hz

The magnetic field energy in an inductor is given by,
E=12Li2
The magnetic energy will be maximum when the current will reach its peak value, i0, and it will be minimum when the current will become zero.



From the above graph of alternating current, we can see that current reduces from maximum to zero in T/4 time, where T is the time period.
So, in this case, T/4 = 5 ms
T=20 ms Frequency, ν = 1T=120×10-3=50 Hz

Page No 329:

Question 6:

(b) 50 Hz

The magnetic field energy in an inductor is given by,
E=12Li2
The magnetic energy will be maximum when the current will reach its peak value, i0, and it will be minimum when the current will become zero.



From the above graph of alternating current, we can see that current reduces from maximum to zero in T/4 time, where T is the time period.
So, in this case, T/4 = 5 ms
T=20 ms Frequency, ν = 1T=120×10-3=50 Hz

Answer:

(d)

The reactance of a series LC circuit is given by,
X=XL-XC=ωL-1ωCX=2πfL-12πfC
The correct relation between the reactance and f is represented by the graph in option (d).
X=0, when XL=XC.

Thus plot d represent the reactance of a series LC combination.

Page No 329:

Question 7:

(d)

The reactance of a series LC circuit is given by,
X=XL-XC=ωL-1ωCX=2πfL-12πfC
The correct relation between the reactance and f is represented by the graph in option (d).
X=0, when XL=XC.

Thus plot d represent the reactance of a series LC combination.

Answer:

(a) 5 Ω

The impedance of the circuit is given by
Z=R2+X2R=4 Ω & X=3 ΩZ=42+32=5 Ω

Page No 329:

Question 8:

(a) 5 Ω

The impedance of the circuit is given by
Z=R2+X2R=4 Ω & X=3 ΩZ=42+32=5 Ω

Answer:

(b) in AC circuits only

When a DC supply is provided to a transformer, there will be no change in flux with time across the coils of the transformer. So, there will be no induced emf in the secondary coil due to changing current in the primary coil. Hence, the transformer cannot operate in DC because of the violation of its working principle.

Page No 329:

Question 9:

(b) in AC circuits only

When a DC supply is provided to a transformer, there will be no change in flux with time across the coils of the transformer. So, there will be no induced emf in the secondary coil due to changing current in the primary coil. Hence, the transformer cannot operate in DC because of the violation of its working principle.

Answer:

(c) i12+i222
Given:
i = i1 cos ωt + i2 sin ωt
The rms value of current is given by,
irms=0Ti2dt0Tdti=i1cos ωt +i2 sin ωtirms=0Ti1cos ωt +i2 sin ωt2dt0Tdtirms=0Ti12cos2ωt +i22 sin2ωt +2i1i2sin ωt cos ωt dt0Tdtirms=0Ti12(cos 2ωt+1)2 +i22 (1-cos 2ωt) 2 +i1i2sin 2ωt  dt0Tdt  [cos2ωt =(cos 2ωt+1)2 , sin2ωt=(1-cos 2ωt) 2 ]

We know that, T = 2π

Integrating the above expression
irms=12i12 02π1dt+02πcos 2ωt dt  + i22 02π1 dt -02π cos 2ωt dt +i1i202π sin 2ωt  dt02πdt 

The following integrals become zero
 02πcos 2ωt dt = 0=02π sin 2ωt  dt
Therefore, it becomes
irms=i12202π1dt+i22202π1dt02πdtirms=i122×2π +i222×2π2πirms =i12+i222

Page No 329:

Question 10:

(c) i12+i222
Given:
i = i1 cos ωt + i2 sin ωt
The rms value of current is given by,
irms=0Ti2dt0Tdti=i1cos ωt +i2 sin ωtirms=0Ti1cos ωt +i2 sin ωt2dt0Tdtirms=0Ti12cos2ωt +i22 sin2ωt +2i1i2sin ωt cos ωt dt0Tdtirms=0Ti12(cos 2ωt+1)2 +i22 (1-cos 2ωt) 2 +i1i2sin 2ωt  dt0Tdt  [cos2ωt =(cos 2ωt+1)2 , sin2ωt=(1-cos 2ωt) 2 ]

We know that, T = 2π

Integrating the above expression
irms=12i12 02π1dt+02πcos 2ωt dt  + i22 02π1 dt -02π cos 2ωt dt +i1i202π sin 2ωt  dt02πdt 

The following integrals become zero
 02πcos 2ωt dt = 0=02π sin 2ωt  dt
Therefore, it becomes
irms=i12202π1dt+i22202π1dt02πdtirms=i122×2π +i222×2π2πirms =i12+i222

Answer:

(d) about 10 A

The rms value of an alternating current is equivalent to the constant current. So, the heating effect produced is actually measured in terms of the rms value, in case of alternating current. The constant current is, thus, equal to the rms value of alternating current, which is given by,
Irms=Ipeak2=142=9.910 A

Page No 329:

Question 11:

(d) about 10 A

The rms value of an alternating current is equivalent to the constant current. So, the heating effect produced is actually measured in terms of the rms value, in case of alternating current. The constant current is, thus, equal to the rms value of alternating current, which is given by,
Irms=Ipeak2=142=9.910 A

Answer:

(a) 2.8 A

The constant current is equal to the rms value of current. So,
Irms= 2.8 A

Page No 329:

Question 1:

(a) 2.8 A

The constant current is equal to the rms value of current. So,
Irms= 2.8 A

Answer:

(a) of the inductor increases

The reactance of an inductor is given by,
XL=ωL
And the reactance of a capacitor is given by,
XC=1ωC
Here, ω = 2πf , where f is the frequency of the source. So, when f increases, ω increases.
XL will increase and XC will decrease.

Page No 329:

Question 2:

(a) of the inductor increases

The reactance of an inductor is given by,
XL=ωL
And the reactance of a capacitor is given by,
XC=1ωC
Here, ω = 2πf , where f is the frequency of the source. So, when f increases, ω increases.
XL will increase and XC will decrease.

Answer:

(a) an inductor and a capacitor
(d) neither an inductor nor a capacitor

(a) The reactance of a circuit containing a capacitor and an inductor is given by,
X=XL-XC=ωL-1ωC
If ωL=1ωC, X = 0. So, the circuit contains an inductor and a capacitor.

(b) For a circuit without a capacitor, reactance is given by,
X'=XL=ωL, which cannot be zero for an AC source.

(c) Similarly, for a circuit without an inductor, reactance is given by,
X''=XC=1ωC, which also cannot be zero.

(d) For a circuit without any capacitor and inductor, reactance, X' = 0 ( L = C = 0)



Page No 330:

Question 3:

(a) an inductor and a capacitor
(d) neither an inductor nor a capacitor

(a) The reactance of a circuit containing a capacitor and an inductor is given by,
X=XL-XC=ωL-1ωC
If ωL=1ωC, X = 0. So, the circuit contains an inductor and a capacitor.

(b) For a circuit without a capacitor, reactance is given by,
X'=XL=ωL, which cannot be zero for an AC source.

(c) Similarly, for a circuit without an inductor, reactance is given by,
X''=XC=1ωC, which also cannot be zero.

(d) For a circuit without any capacitor and inductor, reactance, X' = 0 ( L = C = 0)

Answer:

(a) pure inductor
(b) pure capacitor
(d) combination of an inductor and a capacitor.

For a pure inductive circuit, voltage leads the current by 90°. So, the instantaneous current is zero when the instantaneous voltage is maximum.
Similar is the case with a purely capacitive circuit, in which, current leads the voltage by 90°.

Also, in a circuit containing a combination of inductor and capacitor, the current may lead or lag the voltage by 90°, depending upon whether the voltage across the inductor or the capacitor is greater. Here too, there is a phase difference of 90° between the voltage and current. Hence, the the instantaneous current is zero when the instantaneous voltage is maximum.

Page No 330:

Question 4:

(a) pure inductor
(b) pure capacitor
(d) combination of an inductor and a capacitor.

For a pure inductive circuit, voltage leads the current by 90°. So, the instantaneous current is zero when the instantaneous voltage is maximum.
Similar is the case with a purely capacitive circuit, in which, current leads the voltage by 90°.

Also, in a circuit containing a combination of inductor and capacitor, the current may lead or lag the voltage by 90°, depending upon whether the voltage across the inductor or the capacitor is greater. Here too, there is a phase difference of 90° between the voltage and current. Hence, the the instantaneous current is zero when the instantaneous voltage is maximum.

Answer:

(a) Current
(b) Induced emf in the inductor

For a series L-R circuit, the AC current can be given by,
i=i0sinωt



From the graph, we can see that the average value of current over a cycle is zero.

Since it a series L-R circuit, the phase difference between current and voltage is π2. The AC voltage can be given by,
V=V0cosωt



From the graph, we can see that the average value of voltage over a cycle is also zero.

Joule's heat through the resistor is given by,
Havg=irms2R, which is non zero.

Similarly, magnetic energy stored in the inductor is given by,
Uavg=12Lirms2 , which is also non-zero.

Page No 330:

Question 5:

(a) Current
(b) Induced emf in the inductor

For a series L-R circuit, the AC current can be given by,
i=i0sinωt



From the graph, we can see that the average value of current over a cycle is zero.

Since it a series L-R circuit, the phase difference between current and voltage is π2. The AC voltage can be given by,
V=V0cosωt



From the graph, we can see that the average value of voltage over a cycle is also zero.

Joule's heat through the resistor is given by,
Havg=irms2R, which is non zero.

Similarly, magnetic energy stored in the inductor is given by,
Uavg=12Lirms2 , which is also non-zero.

Answer:

(b) a hot-wire voltmeter

Only a hot-wire voltmeter can be used to measure an AC voltage across a resistor.

Page No 330:

Question 6:

(b) a hot-wire voltmeter

Only a hot-wire voltmeter can be used to measure an AC voltage across a resistor.

Answer:

(a) DC dynamo
(b) AC dynamo

An AC or DC dynamo can be used to convert mechanical energy to electrical energy.
A motor converts electrical energy to mechanical energy and a transformer is used to step up or down the voltage or simply to transfer electrical energy.

Page No 330:

Question 7:

(a) DC dynamo
(b) AC dynamo

An AC or DC dynamo can be used to convert mechanical energy to electrical energy.
A motor converts electrical energy to mechanical energy and a transformer is used to step up or down the voltage or simply to transfer electrical energy.

Answer:

(b) may be 1000 W
(d) may be less than 1000 W

The average power delivered by an AC source is given by,
Pavg=VrmsIrmscosϕ
Given:
Vrms = 100 V
Irms = 10 A
Pavg=1000cosϕ
ϕ-π2,π2
cosϕ0,1
0Pavg1000

Page No 330:

Question 1:

(b) may be 1000 W
(d) may be less than 1000 W

The average power delivered by an AC source is given by,
Pavg=VrmsIrmscosϕ
Given:
Vrms = 100 V
Irms = 10 A
Pavg=1000cosϕ
ϕ-π2,π2
cosϕ0,1
0Pavg1000

Answer:

Frequency of alternating current, f = 50 Hz
Alternation current i is given by,
 i = i0sinωt   ...(1)
Here, i0 = peak value of current
Root mean square value of current irms is given by,
 irms=i02  ...1 
 On substituting the value of the root mean square value of current in place of alternating current in equation (1), we get:
i02=i0sinωt12 = sinωt = sinπ4π4 = ωtt =π4ω = π4×2πf  ω=2πf     = 18f = 18×50    =1400=0.0025 s    =2.5 ms

Page No 330:

Question 2:

Frequency of alternating current, f = 50 Hz
Alternation current i is given by,
 i = i0sinωt   ...(1)
Here, i0 = peak value of current
Root mean square value of current irms is given by,
 irms=i02  ...1 
 On substituting the value of the root mean square value of current in place of alternating current in equation (1), we get:
i02=i0sinωt12 = sinωt = sinπ4π4 = ωtt =π4ω = π4×2πf  ω=2πf     = 18f = 18×50    =1400=0.0025 s    =2.5 ms

Answer:

RMS value of voltage, Erms = 220 V,
Frequency of alternating current, f = 50 Hz
(a) Peak value of voltage E0 is given by,
    E0=Erms 2    ,
where Erms= root mean square value of voltage
   E0=Erms 2 E0=2×220 E0=311.08 V=311 V
(b) Voltage E is given by,
    E=E0sinωt,
where E0 = peak value of voltage
Time taken for the current to reach zero from the rms value = Time taken for the current to reach the rms value from zero
In one complete cycle, current starts from zero and again reaches zero.
So, first we need to find the time taken for the current to reach the rms value from zero.
As E=E02, E02 = E0sin ωtωt = π4t=π4ω=π4×2πf t=π8π50=1400 t= 2.5 ms
Thus, the least possible time in which voltage can change from the rms value to zero is 2.5 ms.

Page No 330:

Question 3:

RMS value of voltage, Erms = 220 V,
Frequency of alternating current, f = 50 Hz
(a) Peak value of voltage E0 is given by,
    E0=Erms 2    ,
where Erms= root mean square value of voltage
   E0=Erms 2 E0=2×220 E0=311.08 V=311 V
(b) Voltage E is given by,
    E=E0sinωt,
where E0 = peak value of voltage
Time taken for the current to reach zero from the rms value = Time taken for the current to reach the rms value from zero
In one complete cycle, current starts from zero and again reaches zero.
So, first we need to find the time taken for the current to reach the rms value from zero.
As E=E02, E02 = E0sin ωtωt = π4t=π4ω=π4×2πf t=π8π50=1400 t= 2.5 ms
Thus, the least possible time in which voltage can change from the rms value to zero is 2.5 ms.

Answer:

Power of the bulb, P = 60 W
Voltage at the bulb, V = 220 V
RMS value of alternating voltage, Erms = 220 V
 P = V2R,
where R = resistance of the bulb
 R=V2P=220×22060      =806.67
Peak value of voltage E0 is given by,
E0=Erms2
  =220×2
  = 311.08
Now, maximum current through the filament i0 is,
i0=E0R
 i0=311.08806.67=0.39 A

Page No 330:

Question 4:

Power of the bulb, P = 60 W
Voltage at the bulb, V = 220 V
RMS value of alternating voltage, Erms = 220 V
 P = V2R,
where R = resistance of the bulb
 R=V2P=220×22060      =806.67
Peak value of voltage E0 is given by,
E0=Erms2
  =220×2
  = 311.08
Now, maximum current through the filament i0 is,
i0=E0R
 i0=311.08806.67=0.39 A

Answer:

Voltage across the electric bulb, E = 12 volts
Let E0 be the peak value of voltage.
We know that heat produced by passing an alternating currenti through a resistor is equal to heat produced by passing a constant currentirms through the same resistor. If R is the resistance of the electric bulb and T is the temperature, then
   i2RT=irms2RTE2R2=Erms2R2E2=E022 E2rms=E202E02=2E2E02=2×122=2×144E0=2×144      =16.97=17 V
Thus , peak value of voltage is 17 V.

Page No 330:

Question 5:

Voltage across the electric bulb, E = 12 volts
Let E0 be the peak value of voltage.
We know that heat produced by passing an alternating currenti through a resistor is equal to heat produced by passing a constant currentirms through the same resistor. If R is the resistance of the electric bulb and T is the temperature, then
   i2RT=irms2RTE2R2=Erms2R2E2=E022 E2rms=E202E02=2E2E02=2×122=2×144E0=2×144      =16.97=17 V
Thus , peak value of voltage is 17 V.

Answer:

Peak power of the resistive coil, P0 =80 W
Time, t = 100 s
RMS value of power Prms is given by,
Prms = P02,
where P0 = Peak value of power
 Prms=P02=40 W
Energy consumed E is given by,
  E = Prms × t
     = 40 × 100
    = 4000 J = 4.0 kJ

Page No 330:

Question 6:

Peak power of the resistive coil, P0 =80 W
Time, t = 100 s
RMS value of power Prms is given by,
Prms = P02,
where P0 = Peak value of power
 Prms=P02=40 W
Energy consumed E is given by,
  E = Prms × t
     = 40 × 100
    = 4000 J = 4.0 kJ

Answer:

Given:
Area of parallel-plate air-capacitor, A = 20 cm2
Separation between the plates, d = 0.1 mm
Dielectric strength of air, E= 3 × 106 V/m
E = Vd,
where V = potential difference across the capacitor
V = Ed
      = 3 × 106 × 0.1 × 10−3
     = 3 × 102 = 300 V
Thus, peak value of voltage is 300 V.
Maximum rms value of voltage Vrms is given by,
Vrms=V02   =3002=212 V

Page No 330:

Question 7:

Given:
Area of parallel-plate air-capacitor, A = 20 cm2
Separation between the plates, d = 0.1 mm
Dielectric strength of air, E= 3 × 106 V/m
E = Vd,
where V = potential difference across the capacitor
V = Ed
      = 3 × 106 × 0.1 × 10−3
     = 3 × 102 = 300 V
Thus, peak value of voltage is 300 V.
Maximum rms value of voltage Vrms is given by,
Vrms=V02   =3002=212 V

Answer:

As per the question,
i = i0e-tτ
We need to find the rms current. So, taking the average of i within the limits 0 to τ and then dividing by the given time period τ, we get:
irms2 = 1τ0τi02 e-2t/τ dtirms2 =i02τ 0τ e-2t/τ dt       =i02τ×-τ2 e-2t/τ0τ     =-i02τ×τ2×e-2-1      =i022(1-1e2)irms=i0e e2-12

Page No 330:

Question 8:

As per the question,
i = i0e-tτ
We need to find the rms current. So, taking the average of i within the limits 0 to τ and then dividing by the given time period τ, we get:
irms2 = 1τ0τi02 e-2t/τ dtirms2 =i02τ 0τ e-2t/τ dt       =i02τ×-τ2 e-2t/τ0τ     =-i02τ×τ2×e-2-1      =i022(1-1e2)irms=i0e e2-12

Answer:

Capacitance of the capacitor, C = 10 μF = 10 × 10−6 F = 10−5 F
Output voltage of the oscillator, ε= (10 V)sinωt
On comparing the output voltage of the oscillator with ε=ε0sinωt, we get:
Peak voltage ε0 = 10 V
For a capacitive circuit,
Reactance, Xc=1ωC
Here, ω = angular frequency
           C = capacitor of capacitance
 Peak current, I0 = ε0Xc
(a) At ω = 10 s−1:
Peak current,
I0 = ε0Xc
    = ε01/ωC=101/10×10-5 A
      = 1 × 10−3 A
(b)  At ω = 100 s−1:
Peak current, I0 = ε01/ωC
I0= 101/100×10-5I0 =10103=1×10-2 A        =0.01 A

(c) At ω = 500 s−1:
Peak current, I0 = ε01ωC
   I0 = ε01/ωCI0 = 101/500×10-5I0 = 10×500×10-5    = 5×10-2 A=0.05 A

(d) At ω = 1000 s−1:
Peak current, I0 = ε01ωC
 I0=101/1000×10-5I0 =10×1000×10-5I0 =10-1 A=0.1 A

Page No 330:

Question 9:

Capacitance of the capacitor, C = 10 μF = 10 × 10−6 F = 10−5 F
Output voltage of the oscillator, ε= (10 V)sinωt
On comparing the output voltage of the oscillator with ε=ε0sinωt, we get:
Peak voltage ε0 = 10 V
For a capacitive circuit,
Reactance, Xc=1ωC
Here, ω = angular frequency
           C = capacitor of capacitance
 Peak current, I0 = ε0Xc
(a) At ω = 10 s−1:
Peak current,
I0 = ε0Xc
    = ε01/ωC=101/10×10-5 A
      = 1 × 10−3 A
(b)  At ω = 100 s−1:
Peak current, I0 = ε01/ωC
I0= 101/100×10-5I0 =10103=1×10-2 A        =0.01 A

(c) At ω = 500 s−1:
Peak current, I0 = ε01ωC
   I0 = ε01/ωCI0 = 101/500×10-5I0 = 10×500×10-5    = 5×10-2 A=0.05 A

(d) At ω = 1000 s−1:
Peak current, I0 = ε01ωC
 I0=101/1000×10-5I0 =10×1000×10-5I0 =10-1 A=0.1 A

Answer:

Given:
Inductance of the coil, = 5.0 mH = 0.005 H
(a) At  ω = 100 s−1:
Reactance of coil XL is given by,
XL = ωL
Here, ω = angular frequency
XL = 100 × 0.005 = 0.5 Ω
Peak current, I0 = 100.5=20 A
(b) At ω = 500 s−1:
   Reactance, XL = 500×51000                   = 2.5 Ω
   Peak current, I0 = 102.5=4 A
(c) ω = 1000 s−1:
Reactance, XL= 1000 ×0.005 = 5 Ω
Peak current, I0 = 105 = 2 A

Page No 330:

Question 10:

Given:
Inductance of the coil, = 5.0 mH = 0.005 H
(a) At  ω = 100 s−1:
Reactance of coil XL is given by,
XL = ωL
Here, ω = angular frequency
XL = 100 × 0.005 = 0.5 Ω
Peak current, I0 = 100.5=20 A
(b) At ω = 500 s−1:
   Reactance, XL = 500×51000                   = 2.5 Ω
   Peak current, I0 = 102.5=4 A
(c) ω = 1000 s−1:
Reactance, XL= 1000 ×0.005 = 5 Ω
Peak current, I0 = 105 = 2 A

Answer:

Given:
Resistance of coil, R = 10 Ω
Inductance of coil, L = 0.4 Henry
Voltage of AC source, Erms = 6.5 V
Frequency of AC source, f30πHz
Reactance of resistance-inductance circuit Z is given by,
 Z = R2+XL2
Here, R = resistance of the circuit
           XL = Reactance of the pure inductive circuit
Z = R2+2πfL2
  = 102+2×π×30π×0.42
Average power consumed in the circuit (P) is given by,
 P = ErmsIrmscosϕ
 cosϕ = RZ, Irms = ErmsZ
 P =6.5×6.5Z×RZ P =6.5×6.5×10R2+ωL22 P =6.5×6.5×10100+576P=6.5×6.5×10676P=0.625=58 W

Page No 330:

Question 11:

Given:
Resistance of coil, R = 10 Ω
Inductance of coil, L = 0.4 Henry
Voltage of AC source, Erms = 6.5 V
Frequency of AC source, f30πHz
Reactance of resistance-inductance circuit Z is given by,
 Z = R2+XL2
Here, R = resistance of the circuit
           XL = Reactance of the pure inductive circuit
Z = R2+2πfL2
  = 102+2×π×30π×0.42
Average power consumed in the circuit (P) is given by,
 P = ErmsIrmscosϕ
 cosϕ = RZ, Irms = ErmsZ
 P =6.5×6.5Z×RZ P =6.5×6.5×10R2+ωL22 P =6.5×6.5×10100+576P=6.5×6.5×10676P=0.625=58 W

Answer:

Given:
Peak voltage of AC source, E0 = 12 V
Angular frequency, ω = 250 π s−1
Resistance of resistor, R = 100 Ω
Energy dissipated as heat H is given by,
H=Erms2RT
Here, Erms = RMS value of voltage
            R = Resistance of the resistor
            T = Temperature
Energy dissipated as heat during t = 0 to t = 1.0 ms,
H=0tdH =E02 sin2 ωtRdt Erms =E0sinωt = 144100010-3sin2 ωt dt = 1.44 010-31-cos 2 ωt2dt =1.442010-3dt+010-3cos 2 ωt dt =0.72 10-3-sin2ωt2ω010-3 = 0.7211000-1500 π = 0.7211000-21000π = π-21000 π×0.72 = 0.0002614 = 2.61×10-4 J

Page No 330:

Question 12:

Given:
Peak voltage of AC source, E0 = 12 V
Angular frequency, ω = 250 π s−1
Resistance of resistor, R = 100 Ω
Energy dissipated as heat H is given by,
H=Erms2RT
Here, Erms = RMS value of voltage
            R = Resistance of the resistor
            T = Temperature
Energy dissipated as heat during t = 0 to t = 1.0 ms,
H=0tdH =E02 sin2 ωtRdt Erms =E0sinωt = 144100010-3sin2 ωt dt = 1.44 010-31-cos 2 ωt2dt =1.442010-3dt+010-3cos 2 ωt dt =0.72 10-3-sin2ωt2ω010-3 = 0.7211000-1500 π = 0.7211000-21000π = π-21000 π×0.72 = 0.0002614 = 2.61×10-4 J

Answer:

Given:
Resistance of the series RC circuit, R = 300 Ω
Capacitance of the series RC circuit, C = 25 μF
Peak value of voltage, ε0 = 50 V
Frequency of the AC source, ν = 50/π Hz            
Capacitive reactance XC is given by,
XC=1ωC
Here, ω = angular frequency of AC source
            C = capacitive reactance of capacitance
  XC=12π×50π×25×10-6 XC=10425 Ω
Net reactance of the series RC circuit Z = R2+XC2
Z = 3002+104252
      = 3002+4002 = 500 Ω
(a) Peak value of current I0 is given by,
    I0= ε0Z I0= 50500 = 0.1 A 

(b) Average power dissipated in the circuit P is given by,
      PεrmsIrms cosÏ•.
      εrms = ε02
       and Irms=I02
   P=E02×I02×RZP=50×0.1×3002×500P =32=1.5 W

Page No 330:

Question 13:

Given:
Resistance of the series RC circuit, R = 300 Ω
Capacitance of the series RC circuit, C = 25 μF
Peak value of voltage, ε0 = 50 V
Frequency of the AC source, ν = 50/π Hz            
Capacitive reactance XC is given by,
XC=1ωC
Here, ω = angular frequency of AC source
            C = capacitive reactance of capacitance
  XC=12π×50π×25×10-6 XC=10425 Ω
Net reactance of the series RC circuit Z = R2+XC2
Z = 3002+104252
      = 3002+4002 = 500 Ω
(a) Peak value of current I0 is given by,
    I0= ε0Z I0= 50500 = 0.1 A 

(b) Average power dissipated in the circuit P is given by,
      PεrmsIrms cosÏ•.
      εrms = ε02
       and Irms=I02
   P=E02×I02×RZP=50×0.1×3002×500P =32=1.5 W

Answer:

Power consumed by the electric bulb, P = 55 W
Voltage at which the bulb is operated, V= 110 V
Voltage of the line, V = 220 V
Frequency of the source, v = 50 Hz
P =V2R,
where R = resistance of electric bulb
  R=V2P
       =110×11055=220 Ω
If L is the inductance of the coil, then total reactance of the circuit (Z) is given by,
 Z = R2+ωL2
   =2202+100πL2    ω=2πf
Here, ω = angular frequency of the circuit
Now, current through the bulb, I=VZ
Voltage drop across the bulb, V = VZ×R
As per question,
110=220×2202202+100πL2110=22022202+100πL2220×2=2202+100πL22202+100πL2=440248400+104π2L2=193600104π2L2=193600-48400L2=145200π2×104       =1.4726L=1.21351.2 H

Page No 330:

Question 14:

Power consumed by the electric bulb, P = 55 W
Voltage at which the bulb is operated, V= 110 V
Voltage of the line, V = 220 V
Frequency of the source, v = 50 Hz
P =V2R,
where R = resistance of electric bulb
  R=V2P
       =110×11055=220 Ω
If L is the inductance of the coil, then total reactance of the circuit (Z) is given by,
 Z = R2+ωL2
   =2202+100πL2    ω=2πf
Here, ω = angular frequency of the circuit
Now, current through the bulb, I=VZ
Voltage drop across the bulb, V = VZ×R
As per question,
110=220×2202202+100πL2110=22022202+100πL2220×2=2202+100πL22202+100πL2=440248400+104π2L2=193600104π2L2=193600-48400L2=145200π2×104       =1.4726L=1.21351.2 H

Answer:

Given:
Resistance in series LCR circuit, R = 300 Ω
Capacitance in series LCR circuit, C = 20 μF= 20 × 10−6 F
Inductance in series LCR circuit, L = 1 Henry
RMS value of voltage, εrms   = 50 V
Frequency of source, f = 50/π Hz
Reactance of the inductor (XL) is given by,
 XL= ωL = 2πfL
  XL = 2×π×50π×1  = 100 Ω
Reactance of the capacitance XC is given by,
XC=1ωC = 12πfC
XC = 12π×50π×20×10-6
XC  = 500Ω                                                  
(a) Impedance of  an LCR circuit Z is given by,
Z=R2+XC-XL2      Z  =3002+500-1002  Z  =3002+4002 Z =500
RMS value of current Irms is given by,
Irms=εrmsZ
Irms = 50500
Irms = 0.1 A
(b) Potential across the capacitor VC is given by,
VC = Irms × XC
VC = 0.1 × 500 = 50 V
Potential difference across the resistor VR is given by,
VR = Irms × R
 VR = 0.1 × 300 = 30 V
Potential difference across the inductor VL is given by,
VL = Irms × XL
 VL= 0.1 × 100 = 10 V
R.M.S potential = 50 V
Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
Sum of the potential drops > RMS potential applied

Page No 330:

Question 15:

Given:
Resistance in series LCR circuit, R = 300 Ω
Capacitance in series LCR circuit, C = 20 μF= 20 × 10−6 F
Inductance in series LCR circuit, L = 1 Henry
RMS value of voltage, εrms   = 50 V
Frequency of source, f = 50/π Hz
Reactance of the inductor (XL) is given by,
 XL= ωL = 2πfL
  XL = 2×π×50π×1  = 100 Ω
Reactance of the capacitance XC is given by,
XC=1ωC = 12πfC
XC = 12π×50π×20×10-6
XC  = 500Ω                                                  
(a) Impedance of  an LCR circuit Z is given by,
Z=R2+XC-XL2      Z  =3002+500-1002  Z  =3002+4002 Z =500
RMS value of current Irms is given by,
Irms=εrmsZ
Irms = 50500
Irms = 0.1 A
(b) Potential across the capacitor VC is given by,
VC = Irms × XC
VC = 0.1 × 500 = 50 V
Potential difference across the resistor VR is given by,
VR = Irms × R
 VR = 0.1 × 300 = 30 V
Potential difference across the inductor VL is given by,
VL = Irms × XL
 VL= 0.1 × 100 = 10 V
R.M.S potential = 50 V
Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
Sum of the potential drops > RMS potential applied

Answer:

Given:
Resistance in the LCR circuit, R = 300 Ω
Capacitance in the LCR circuit, C = 20 μF = 20 × 10−6 F
Inductance in the LCR circuit, L = 1 henry
Net impedance of the LCR circuit, Z = 500 ohm
RMS value of voltage, εrms = 50 V
RMS value of current, Irms = 0.1 A
Peak current I0 is given by,
I0=Erms2Z=50×1.414500=0.1414 A
Electrical energy stored in capacitorUC is given by,
 UC=12 CV2
 UC=12×20×10-6×50×50UC=25×10-3 J=25 mJ
Magnetic field energy stored in the coil UL is given by,
    UL=12LI02
  UL=12×1×0.14142UL 5×10-3 JUL=5 mJ

Page No 330:

Question 16:

Given:
Resistance in the LCR circuit, R = 300 Ω
Capacitance in the LCR circuit, C = 20 μF = 20 × 10−6 F
Inductance in the LCR circuit, L = 1 henry
Net impedance of the LCR circuit, Z = 500 ohm
RMS value of voltage, εrms = 50 V
RMS value of current, Irms = 0.1 A
Peak current I0 is given by,
I0=Erms2Z=50×1.414500=0.1414 A
Electrical energy stored in capacitorUC is given by,
 UC=12 CV2
 UC=12×20×10-6×50×50UC=25×10-3 J=25 mJ
Magnetic field energy stored in the coil UL is given by,
    UL=12LI02
  UL=12×1×0.14142UL 5×10-3 JUL=5 mJ

Answer:

Given:
Inductance of inductor, L = 2.0 H
Capacitance of capacitor, C = 18 μF
Resistance of resistor, R = 10 kΩ
Voltage of AC source, E = 20 V
(a) In an LCR circuit, current is maximum when reactance is minimum, which occurs at resonance, i.e. when capacitive reactance becomes equal to the inductive reactance,i.e.
  XL = XC
ωL=1ωCω2=1LC=12×18×10-6ω2=10636ω=10362πf=1036f=10006×2π=26.539 Hz     =27 Hz

(b) At resonance, reactance is minimum.
     Minimum Reactance, Z = R
     Maximum current (I) is given by,
       I=ERI=2010×103
   I=2A103=2 mA     

Page No 330:

Question 17:

Given:
Inductance of inductor, L = 2.0 H
Capacitance of capacitor, C = 18 μF
Resistance of resistor, R = 10 kΩ
Voltage of AC source, E = 20 V
(a) In an LCR circuit, current is maximum when reactance is minimum, which occurs at resonance, i.e. when capacitive reactance becomes equal to the inductive reactance,i.e.
  XL = XC
ωL=1ωCω2=1LC=12×18×10-6ω2=10636ω=10362πf=1036f=10006×2π=26.539 Hz     =27 Hz

(b) At resonance, reactance is minimum.
     Minimum Reactance, Z = R
     Maximum current (I) is given by,
       I=ERI=2010×103
   I=2A103=2 mA     

Answer:

RMS value of voltage, Erms = 24 V
Internal resistance of battery, r = 4 Ω
RMS value of current, Irms = 6 A
Reactance R is given by,
  R=EIrmsR=246=4 Ω
Let R' be the total resistance of the circuit. Then,
   R' = R + r
 R
' 4 Ω +  4 Ω    
 R' = 8 Ω

Current, I = 128 A
                = 1.5 A

Page No 330:

Question 18:

RMS value of voltage, Erms = 24 V
Internal resistance of battery, r = 4 Ω
RMS value of current, Irms = 6 A
Reactance R is given by,
  R=EIrmsR=246=4 Ω
Let R' be the total resistance of the circuit. Then,
   R' = R + r
 R
' 4 Ω +  4 Ω    
 R' = 8 Ω

Current, I = 128 A
                = 1.5 A

Answer:

Here,
Input voltage to the filter, Vi = 10 × 10−3V
Resistance of the circuit, R = 1 × 103 Ω
Capacitance of the circuit, C = 10 × 10−9 F
(a) When frequency, f = 10 kHz
  A low pass filter consists of resistance and capacitance. Voltage across the capacitor is taken as the output.
     Capacitive reactance XC is given by,
       XC=1ωC=12πfC
   XC=12π×10×103×10×10-9XC=12π×10-4XC   =1042π=5000π Ω
  Net impedence of the resistance-capacitance circuit (Z) is given by,
    Z=R2+XC2Z=1+103+5000/π2 Z=106+5000/π2
 Current (I0) is given by,
    I0=ViZI0=10×10-3106+5000/π2
Output across the capacitor V0 is given by,
       V0=102105+50/π2×500π    V0 =1.6124 V=1.6 mV
(b)When frequency, f = 1 MHz = 1×106 Hz
Capacitive reactance XC is given by,
Xc=1ωCXC=12πfCXC=12π×106×10-9×10XC =12π×10-2XC=1002πXC=500π ΩTotal impedence Z =R2+XC2Z =1032+50/π2Current (I0) =V1ZI0 =10×10-3106+50/π2Output voltage V0 =I0XCV0=10-2105+50/π2×50πV0=0.16 mV

(c) When frequency, f = 10 MHz = 10 × 106 Hz = 107 Hz
       Capacitive reactance XC is given by,
       Xc=1ωCXC=12πfCXC=12π×107×10×10-9XC=5π ΩImpedence Z =R2+Xc2Z=1032+5/π2Current I0=V1ZI0=10×10-3106+5/π2V0=I0XCV0=10-2106+5/π2×5πVo=16 μV



Page No 331:

Question 19:

Here,
Input voltage to the filter, Vi = 10 × 10−3V
Resistance of the circuit, R = 1 × 103 Ω
Capacitance of the circuit, C = 10 × 10−9 F
(a) When frequency, f = 10 kHz
  A low pass filter consists of resistance and capacitance. Voltage across the capacitor is taken as the output.
     Capacitive reactance XC is given by,
       XC=1ωC=12πfC
   XC=12π×10×103×10×10-9XC=12π×10-4XC   =1042π=5000π Ω
  Net impedence of the resistance-capacitance circuit (Z) is given by,
    Z=R2+XC2Z=1+103+5000/π2 Z=106+5000/π2
 Current (I0) is given by,
    I0=ViZI0=10×10-3106+5000/π2
Output across the capacitor V0 is given by,
       V0=102105+50/π2×500π    V0 =1.6124 V=1.6 mV
(b)When frequency, f = 1 MHz = 1×106 Hz
Capacitive reactance XC is given by,
Xc=1ωCXC=12πfCXC=12π×106×10-9×10XC =12π×10-2XC=1002πXC=500π ΩTotal impedence Z =R2+XC2Z =1032+50/π2Current (I0) =V1ZI0 =10×10-3106+50/π2Output voltage V0 =I0XCV0=10-2105+50/π2×50πV0=0.16 mV

(c) When frequency, f = 10 MHz = 10 × 106 Hz = 107 Hz
       Capacitive reactance XC is given by,
       Xc=1ωCXC=12πfCXC=12π×107×10×10-9XC=5π ΩImpedence Z =R2+Xc2Z=1032+5/π2Current I0=V1ZI0=10×10-3106+5/π2V0=I0XCV0=10-2106+5/π2×5πVo=16 μV

Answer:

A transformer works on the principle of electromagnetic induction, which is only possible in case of AC.
Hence, when DC (zero frequency) is supplied to it, the primary coil blocks the current supplied to it. Thus, the induced current in the secondary coil is zero. So, the output voltage will be zero.



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