Question

A cell of emf (E) and internal resistance (r), is connected to resistance R, the maximum power loss in the resistance R will be:

• A. $\frac{E^2}{R}$
• B. $\frac{E^2}{2R}$
• C. $\frac{E^2}{3R}$
• D. $\frac{E^2}{4R}$

Answer 1

Answer: (D)

$\frac{E^2}{4R}$

Answer 2

The current in the circuit is $I = \frac{E}{R+r}$. The power dissipated in R is $P = I^2 R = \frac{E^2 R}{(R+r)^2}$. For maximum power transfer, the external resistance R must be equal to the internal resistance r ($R=r$). Substituting $R=r$ into the power equation gives the maximum power: $P_{max} = \frac{E^2 r}{(r+r)^2} = \frac{E^2 r}{(2r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}$. Since $R=r$ at maximum power, this can also be written as $\frac{E^2}{4R}$.