Q.
A cell of internal resistance $r$ is connected to a load of resistance $R$. Energy is dissipated in the load, but some thermal energy is also wasted in the cell. The efficiency of such an arrangement is found from the expression
$\frac{energy\ dissipated\ in\ the\ load}{energy\ dissipatd\ in\ the\ compete\ circuit}.$
Which of the following gives the efficiency in this case?
A) $\frac{r}{R}$
B) $\frac{R}{r}$
C) $\frac{r}{R + r}$
D) $\frac{R}{R + r}$
Explanation
(d) Assuming current $I$ flows through the circuit
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Energy dissipated in load $= I^{2}R$
Energy dissipated in the compete circuit $= I^{2}(r + R)$
$\therefore$ The efficiency $= \frac{I^{2}R}{I^{2}(R + r)} = \frac{R}{R + r}$
Relation between resistance, mass and cross-sectional area
$R = p\frac{l \times \alpha}{\alpha \times \alpha}\ or\ R = \rho\frac{V}{\alpha^{2}}$
$R = \rho\frac{m}{d\alpha^{2}} = \left\lbrack \ \frac{\rho}{d} \right\rbrack\frac{m}{\alpha^{2}}$
$I \propto \frac{m}{\alpha^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.