QT P1 9

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The power dissipated in resistor is 15W. The reading of ammeter and voltmeter are 500mA and 10V respectively. Ammeter, voltmeter and battery are ideal

(A) the resistance R1 is 20 ohm

(B) the resistance R3 is 15 ohm

(D) power supplied by battery is 22.5 W

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5 Answers

UTTARA ·2010-02-28 12:58:50

1 2 3

Options r correct

But what abt 4th??

This was my doubt

ANS GIVEN : 1 2 3 4

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rahul nair ·2010-02-28 21:39:51

Power supplied =V²/R

Eff resistance=(R1+R2) in parallel with R3
=10ohms
Power=22.5W

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UTTARA ·2010-02-28 21:43:40

V remains same in parallel

So power = 15 x 15 / 15 no??

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rahul nair ·2010-02-28 21:47:45

V remains same in parallel

V across R3=15
Pow.=15*15/15=15

V across R1+R2=15
Pow=15*15/30=7.5

total pow=p1+p2=22.5

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UTTARA ·2010-02-28 21:53:41

Ok Thanks

Agreed with the given answer

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QT P1 9

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