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Question

A battery of e.m.f. 15 V and internal resistance 2 Ω is connected to two resistors of resistances 4 Ω and 6 Ωjoined in series. Find the electrical energy spent per minute in 6 Ω resistor.

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Solution

Given,
E.m.f.of battery , v=15v
Internal resistance of battery,RB=2Ω
Resistance given in circuit,R1=4Ω
R2=6Ω
i) When resistors re connected in series
Equivalent resistance,R=RB+R1+R1+R2=12Ω
Current the circuit,I=1512=1.25A
Now voltage across resistors R2,V2=IR=1.25×6
V2=7.50V
Time,t=1 min=60sec
Energy across R2,E=V2tR=(7.5)2×606
E=562.5J

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