When resistance of 30 - is connected to a battery- the current in the circuit is 0-4 A- if this resistance is replaced by 10 - resistance- the current is 0-8 A- Then e-m-f- and the internal resistance of thebattery areA- 24 V - 32 -B- 10 V - 16 -C- 3 V - 20 -D- 16 V - 10 -

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Standard XII

Physics

RLC Circuit

When resistan...

Question

When resistance of 30 $Ω$ is connected to a battery, the current in the circuit is 0.4 A. if this resistance is replaced by 10 $Ω$ resistance, the current is 0.8 A. Then e.m.f. and the internal resistance of thebattery are

10 V, 16 Ω

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3 V, 20 Ω

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16 V, 10 Ω

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24 V, 32 Ω

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Solution

The correct option is C $16 V, 10 Ω$
$E = I_{1} [R_{1} + r] = I_{2} [R_{2} + r] 0.4 [30 + r] = 0.8 [10 + r] 30 + r = 20 + 2 r r = 10 E = 0.4 [30 + 10] E = 16 V$

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When resistance of 30Ω is connected to a battery, the current in the circuit is 0.4 A. if this resistance is replaced by 10 Ω resistance, the current is 0.8 A. Then e.m.f. and the internal resistance of thebattery are

The correct option is C 16V,10ΩE=I1[R1+r]=I2[R2+r]0.4[30+r]=0.8[10+r]30+r=20+2rr=10E=0.4[30+10]E=16V

When resistance of 30 $Ω$ is connected to a battery, the current in the circuit is 0.4 A. if this resistance is replaced by 10 $Ω$ resistance, the current is 0.8 A. Then e.m.f. and the internal resistance of thebattery are

The correct option is C $16 V, 10 Ω$
$E = I_{1} [R_{1} + r] = I_{2} [R_{2} + r] 0.4 [30 + r] = 0.8 [10 + r] 30 + r = 20 + 2 r r = 10 E = 0.4 [30 + 10] E = 16 V$