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
A
10V,16Ω
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B
3V,20Ω
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C
16V,10Ω
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D
24V,32Ω
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Solution
The correct option is C16V,10Ω E=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