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

Answer 1

Answer: (A)

The thermal power generated in resistance R will be highest at $R = \frac{R_1 R_2}{R_1 + R_2}$. 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 circuit consists of two batteries with EMFs $E_1, E_2$ and internal resistances $R_1, R_2$ connected in parallel with an external resistor R. This configuration can be simplified by finding the equivalent EMF ($E_{eq}$) and equivalent internal resistance ($R_{eq}$) of the parallel combination of the two batteries.

The equivalent EMF of two sources in parallel with the same polarity is given by: $E_{eq} = \frac{E_1/R_1 + E_2/R_2}{1/R_1 + 1/R_2} = \frac{E_1 R_2 + E_2 R_1}{R_1 + R_2}$

The equivalent internal resistance of two resistances in parallel is given by: $R_{eq} = \frac{1}{1/R_1 + 1/R_2} = \frac{R_1 R_2}{R_1 + R_2}$

The circuit is now equivalent to a single source with EMF $E_{eq}$ and internal resistance $R_{eq}$ connected in series with the external resistor R. The current flowing through R is given by: $I = \frac{E_{eq}}{R_{eq} + R}$

The thermal power generated in R is $P_R = I^2 R$: $P_R = \left(\frac{E_{eq}}{R_{eq} + R}\right)^2 R = \frac{E_{eq}^2 R}{(R_{eq} + R)^2}$

To find the value of R that maximizes $P_R$, we can use the maximum power transfer theorem, which states that maximum power is transferred when the load resistance equals the internal resistance of the source. In this case, the external resistor R is the load, and $R_{eq}$ is the internal resistance of the equivalent source. Therefore, maximum power is generated in R when: $R = R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$

To find the maximum power, we substitute $R = R_{eq}$ into the expression for the current: $I_{max} = \frac{E_{eq}}{R_{eq} + R_{eq}} = \frac{E_{eq}}{2 R_{eq}}$ $I_{max} = \frac{\frac{E_1 R_2 + E_2 R_1}{R_1 + R_2}}{2 \left(\frac{R_1 R_2}{R_1 + R_2}\right)} = \frac{E_1 R_2 + E_2 R_1}{2 R_1 R_2}$

The maximum thermal power generated in R is $P_{max} = I_{max}^2 R_{eq}$: $P_{max} = \left(\frac{E_1 R_2 + E_2 R_1}{2 R_1 R_2}\right)^2 \left(\frac{R_1 R_2}{R_1 + R_2}\right)$ $P_{max} = \frac{(E_1 R_2 + E_2 R_1)^2}{4 R_1^2 R_2^2} \cdot \frac{R_1 R_2}{R_1 + R_2}$ $P_{max} = \frac{(E_1 R_2 + E_2 R_1)^2}{4 R_1 R_2 (R_1 + R_2)}$