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Two lamps,one rated 100W at 220V and the other 60W at 220v are connected in parallel to a 220v supply.What current is drawn from the supply line?

in this

P1=100 W , V = 220 V , P2= 60 W (GIVEN)

I1 = ? I2 = ?

P1= VI1

. : I1=P1/ V

= 100/220 = 0.46A

SIMILIARLY ,

I2=P2/ V = 60/220

= 0.27A

TOTAL CURRENT (I) = I1 + I2

= 0.46 + 0.27

= 0.73A

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View Full Answer

we know that,

power (P)=potential differenc(V) . current(I)

hence P1/ V=I1

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  • -22
in case of first lamp, P=100W, P.D. =220V
I=P/V=100/220=0.45​A

in case of SECOND lamp P=60W, P.D.= 220V
I=P/V=60/220=0.27A

TOTAL CURRENT = 0.45+0.27=0.72 A
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As we know that P=V2/R
then,R=V2/P
resistance of first lamp is     R=(220)2/100= 484 ohm
resistance of second lamp is     R=(220)2/60= 806.7 ohm
since they are connected in parallel combination then,
1/Req=1/R1+1/R2
1/Req=1/484 + 1/806.7
Req= 302.6 ohm
therefore, total current drawn is I=V/R
I=0.73A
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Can we add power p=p1+p2 in this case
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P=VI
I=P/V
​I1=100/220=0.45A
I2=60/220=0.27A
TOTAL current=0.72A
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Solution:

Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be same that is 220 V, Current drawn by the bulb of rating 100 W is given by

I=P/V

?=100/220=0.4545...A

Similarly, current drawn by the bulb of rating 60 W is given by

I=P/V

?=60/220=0.2727...A

??the total current drawn from the line =0.4545+0.2727=0.727A
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