Science And Technology Solutions Solutions for Class 9 Science Chapter 3 - Current Electricity

Answer:

A. The appliances are connected in parallel.

B. The potential difference across each appliance should be same as they are connected in parallel. The potential difference between live and neutral wire in domestic wiring is maintained at 220 V in India. Hence, the potential difference across each appliance should be 220 V.

C. No, the current passing through each appliance is not same. From Ohm's law, the current flowing through each appliance is given as
$I=\frac{V}{R}$
Since, potential difference across each appliance is same, thus
$I\propto \frac{1}{R}$
Now, each appliance has different value of resistance. Thus, the current flowing through them will be different. The one with greater value of resistance will receive lesser amount of current and the one with less value of resistance will receive more current.

D. Domestic appliance are connected in parallel because this connection assures that even if one or more appliances get faulty or stops working, the working of others will remain unaffected.

E. The working of other appliances will remain unaffected even if the T.V. stops working. The current through T.V. will stop but the other appliances will receive the approximately the same amount of current as before. This is because in parallel connection, the current divides itself in the number of electrical branches present. If one of the branches becomes open circuited due to some fault, the remaining branches will still be forming complete circuit for the flow of current.

Answer:

Ohm's law can be proved with the help of above circuit.

Answer:

When Umesh connects the bulbs one at a time, the filaments of the bulbs get burnt. This means the heat energy generated in the filament is beyond what it can tolerate. Hence, we need to reduce the heat generated across the filament. Now, the heat generated in a resistor is proportional to the amount of current flowing through it. Thus, we need to reduce the current flowing through the bulbs. 

A. If bulbs are connected in series, the effective resistance of the circuit increases and hence current through the bulbs decreases. In this way, both the bulbs can be saved from burning.

B. Following are the characteristics of series connection:

• The effective resistance of the circuit is the sum of individual resistance of the resistors present in the circuit. 
• The effective resistance of the series circuit is always larger than the greatest resistance in the circuit.
• Current through various electrical components connected in series is same.
• The source voltage gets divided across the electrical components connected in series.

Answer:

a. From Ohm's law, we have
$R=\frac{V}{I}$

Answer:

Answer:

Resistivity of a conductor, $\rho =R\frac{A}{L}$
Here, L = x, R = r and A = a
Therefore, $\rho =r\frac{a}{x}$
The unit of resistivity is ohm-metre ($\mathrm{\Omega } \mathrm{m}$).

Answer:

a. The circuit for this case is shown below.

Here, R₁ and R₂ are in parallel. Their effective resistance is
$R\text{'}=\frac{R_1\times R_2}{R_1+R_2}$
Also, R₄ is in parallel with a conductor of 0 resistance. Their equivalent resistance is therefore 0. The given circuit thus becomes as shown below.

Now, R' and R₃ are in series. Thus, the equivalent resistance of the circuit is
$R_{\mathrm{eq}}=\frac{R_1\times R_2}{R_1+R_2}+R_3=\frac{R_1{R}_2+R_1{R}_3+R_3{R}_2}{R_1+R_2}$

Hence, the current flowing in the circuit is
$I=\frac{V}{\frac{R_1{R}_2+R_1{R}_3+R_3{R}_2}{R_1+R_2}}=\frac{V(R_1+R_2)}{R_1{R}_2+R_1{R}_3+R_3{R}_2}$
where, V is the potential difference of the battery.

b. The circuit for this case is shown below.

Here, R₁, R₄ and R₃ are in series. Therefore, the equivalent resistance of the circuit is
$R_{\mathrm{eq}}=R_1+R_2+R_3$

Hence, the current flowing in the circuit is
$I=\frac{V}{R_{\mathrm{eq}}}=\frac{V}{R_1+R_2+R_3}$
where, V is the potential difference of the battery.

c. The circuit for this case is shown below.

Here, R₁ and R₂ are in parallel. Their effective resistance is
$R\text{'}=\frac{R_1\times R_2}{R_1+R_2}$

Now, R₃, R' and R₄ are in series.  Thus, the equivalent resistance of the circuit is
$R_{\mathrm{eq}}=\frac{R_1\times R_2}{R_1+R_2}+R_3+R_4=\frac{R_1{R}_2+R_1{R}_3+R_3{R}_2+R_1{R}_4+R_4{R}_2}{R_1+R_2}$

Hence, the current flowing in the circuit is
$I=\frac{V}{\frac{R_1{R}_2+R_1{R}_3+R_3{R}_2+R_1{R}_4+R_4{R}_2}{R_1+R_2}}=\frac{V(R_1+R_2)}{R_1{R}_2+R_1{R}_3+R_3{R}_2+R_1{R}_4+R_4{R}_2}$
where, V is the potential difference of the battery.

Answer:

a. In this case, x₁ , x₂ and x₃ are connected in series.

b.  In this case, x₁ , x₂ and x₃ are connected in series.

c.  In this case, x₁ , x₂ and x₃ are connected in parallel.

d. In this case, x₁ , x₂ and x₃ are connected in parallel.

e. In this case, x₁ , x₂ and x₃ are connected in series.

f. In this case, x₁ , x₂ and x₃ are connected in parallel.

Answer:

A. Resistance of a wire is directly proportional to its length. Let R₁ and R₂ be the resistances of the nichrome wire at 100 cm (1 m) and 70 cm, respectively. Therefore,
$$\begin{gathered} R_1\propto 100 .....\left(\mathrm{i}\right) \\ \mathrm{and} \\ {R}_2\propto 70 .....\left(\mathrm{ii}\right) \\ \mathrm{Dividing} \left(\mathrm{ii}\right) \mathrm{by}\hspace{0.17em}\left(\mathrm{i}\right), \mathrm{we} \mathrm{get} \\ \frac{R_2}{R_1}=\frac{70}{100} \\ \Rightarrow R_{\mathit{2}}=\frac{70}{100}\times R_1 \end{gathered}$$
Now, R₁ = 6 $\mathrm{\Omega }$
$\Rightarrow R_{\mathit{2}}=\frac{70}{100}\times 6=4.2 \mathrm{\Omega }$

B. Let the two resistances be R₁ and R₂. Therefore,
R₁ + R₂ = 80 $\mathrm{\Omega }$
or, R₁ = 80 $-R_2$   .....(i)
and 
$$\begin{gathered} \frac{1}{R_1}+\frac{{\displaystyle 1}}{{\displaystyle R_2}}=\frac{{\displaystyle 1}}{20} \\ \Rightarrow R_1{R}_2=20(R_1+R_2) \\ \mathrm{As}, R_1=80-R_2 \\ \Rightarrow (80-R_2)R_2=20(80 -R_2+R_2) \\ 80R_2^2-R_2=1600 \\ {R}_2^2-80R_2+1600=0 \\ (R_2-40)(R_2-40)=0 \\ \Rightarrow R_2=40 \mathrm{\Omega } \\ \mathrm{From} \left(\mathrm{i}\right), \mathrm{we} \mathrm{get} \\ {R}_1=80-40=40 \mathrm{\Omega } \end{gathered}$$

c. Current flowing through the conducting wire is
 $$\begin{gathered} I=\frac{\mathrm{Charge}\left(Q\right)}{\mathrm{Time} \left(t\right)} \\ \Rightarrow I=\frac{420}{5\times 60}=1.4 \mathrm{A} \end{gathered}$$
Hence, the current flowing through the wire is 1.4 A.