Current Electricity

JEE Main level, single correct, medium

316 of 380

Question 316 of 380

A battery with EMF EE and internal resistance rr is connected across a resistance RR. The power consumption in RR will be maximum when :

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Answer: (C) R=rR=r

Power consumed in external resistance R is P=E2R(R+r)2P = \frac{E^2 R}{(R + r)^2}.
To find maximum, differentiate with respect to R:
dPdR=E2(R+r)2−2R(R+r)(R+r)4=E2r−R(R+r)3\frac{dP}{dR} = E^2 \frac{(R + r)^2 - 2R(R + r)}{(R + r)^4} = E^2 \frac{r - R}{(R + r)^3}.
dPdR=0\frac{dP}{dR} = 0 gives R=rR = r.
This is the maximum power transfer theorem: maximum power is delivered when the external load resistance equals the source internal resistance.

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