Question

Consider a circuit where a cell of emf \( E_0 \) and internal resistance \( r \) is connected across the terminal A and B as shown in the figure. The value of \( R \) for which the power generated in the circuit is maximum, is given by:

• A. \( R = r \)
• B. \( R = 2r \)
• C. \( R = 3r \)
• D. \( R = \frac{r}{3} \)

Hint:

The maximum power in a circuit occurs when the external resistance equals the internal resistance, i.e., R = r. For circuits involving both internal and external resistances, adjust accordingly for maximum power.

Answer 1

The Correct Option is C

Step 1: The total resistance in the circuit is the sum of the internal resistance \( r \) and the external resistance \( R \).

Step 2: The power generated in the circuit is given by:

\[ P = \frac{E_0^2}{R_{\text{total}}} \times R = \frac{E_0^2}{(R + r)^2} \times R, \]

where \( R_{\text{total}} = R + r \) is the total resistance in the circuit.

Step 3: To find the value of \( R \) that maximizes the power, we take the derivative of the power with respect to \( R \) and set it equal to zero:

\[ \frac{dP}{dR} = \frac{E_0^2 (R + r) - E_0^2 R \times 2(R + r)}{(R + r)^4} = 0. \]

Simplifying, we find:

\[ (R + r) - 2R = 0, \]

\[ R = 3r. \]

Thus, the value of \( R \) for which the power is maximum is \( R = 3r \).