Question

In the circuit shown in figure, the emfs of batteries are $E_1$ and $E_2$ which have internal resistances $R_1$ and $R_2$. At what value of the resistance $R$ will the thermal power generated in it be the highest? What it is?

• A. The thermal power generated in resistance $R$ is highest when $R = \frac{R_1 R_2}{R_1 + R_2}$.
• B. The thermal power generated in resistance $R$ is highest when $R = R_1 + R_2$.
• C. The thermal power generated in resistance $R$ is highest when $R = \frac{E_1 R_2 + E_2 R_1}{R_1 + R_2}$.
• D. The thermal power generated in resistance $R$ is highest when $R = \frac{E_1}{R_1} + \frac{E_2}{R_2}$.

Answer 1

Answer: (A)

The thermal power generated in resistance $R$ is highest when $R = \frac{R_1 R_2}{R_1 + R_2}$. The highest thermal power generated is $P_{max} = \frac{(E_1 R_2 + E_2 R_1)^2}{4 R_1 R_2 (R_1 + R_2)}$.

Answer 2

The thermal power generated in resistor $R$ is given by $P = \frac{V^2}{R}$, where $V$ is the voltage across $R$. By applying Kirchhoff's laws, we can express $V$ in terms of the circuit parameters. The currents through the branches with batteries are $I_1 = \frac{E_1 - V}{R_1}$ and $I_2 = \frac{E_2 - V}{R_2}$. By Kirchhoff's current law at the junction, $I_1 + I_2 = \frac{V}{R}$. Substituting the expressions for $I_1$ and $I_2$: $\frac{E_1 - V}{R_1} + \frac{E_2 - V}{R_2} = \frac{V}{R}$ $\frac{E_1}{R_1} + \frac{E_2}{R_2} = V \left( \frac{1}{R} + \frac{1}{R_1} + \frac{1}{R_2} \right)$ Let $S = \frac{E_1}{R_1} + \frac{E_2}{R_2}$ and $G = \frac{1}{R_1} + \frac{1}{R_2}$. Then, $S = V \left( \frac{1}{R} + G \right)$, which gives $V = \frac{S}{\frac{1}{R} + G} = \frac{SR}{1 + GR}$. The power is $P(R) = \frac{V^2}{R} = \frac{1}{R} \left( \frac{SR}{1 + GR} \right)^2 = \frac{S^2 R}{(1 + GR)^2}$. To maximize $P(R)$, we find $\frac{dP}{dR}$ and set it to zero: $\frac{dP}{dR} = S^2 \frac{(1+GR)^2 - R \cdot 2(1+GR)G}{(1+GR)^4} = S^2 \frac{1 - GR}{(1+GR)^3}$ Setting $\frac{dP}{dR} = 0$ gives $1 - GR = 0$, so $R = \frac{1}{G}$. Substituting $G = \frac{1}{R_1} + \frac{1}{R_2} = \frac{R_1 + R_2}{R_1 R_2}$, we get: $R = \frac{1}{\frac{R_1 + R_2}{R_1 R_2}} = \frac{R_1 R_2}{R_1 + R_2}$ The maximum power is obtained by substituting $R = \frac{1}{G}$ into the power equation: $P_{max} = \frac{S^2 (1/G)}{(1 + G(1/G))^2} = \frac{S^2/G}{2^2} = \frac{S^2}{4G}$ Substituting $S = \frac{E_1}{R_1} + \frac{E_2}{R_2} = \frac{E_1 R_2 + E_2 R_1}{R_1 R_2}$ and $G = \frac{R_1 + R_2}{R_1 R_2}$: $P_{max} = \frac{1}{4 \left( \frac{R_1 + R_2}{R_1 R_2} \right)} \left( \frac{E_1 R_2 + E_2 R_1}{R_1 R_2} \right)^2 = \frac{R_1 R_2}{4 (R_1 + R_2)} \frac{(E_1 R_2 + E_2 R_1)^2}{(R_1 R_2)^2} = \frac{(E_1 R_2 + E_2 R_1)^2}{4 R_1 R_2 (R_1 + R_2)}$