Question

In a circuit, there is a battery of emf \(E\) and internal resistance \(r\), connected to an external load resistance \(R\) as shown. Find the value of \(R\) so that maximum power is dissipated across \(R\).

• A. \( R = r \)
• B. \( R = \dfrac{r}{2} \)
• C. \( R = \sqrt{2}\,r \)
• D. \( R = 2r \)

Hint:

In circuits with internal resistance, maximum power transfer occurs when: \[ R_{\text{load}} = r_{\text{internal}} \]

Answer 1

The Correct Option is A

Concept: This question is based on the Maximum Power Transfer Theorem. According to this theorem, maximum power is delivered to the external load when the load resistance is equal to the internal resistance of the source.
Step 1: Expression for current in the circuit The total resistance of the circuit is: \[ R_{\text{total}} = R + r \] Hence, the current flowing through the circuit is: \[ I = \frac{E}{R + r} \]
Step 2: Expression for power dissipated across load resistance \(R\) Power dissipated across \(R\) is: \[ P = I^2 R \] Substituting the value of current: \[ P = \left(\frac{E}{R + r}\right)^2 R \] \[ P = \frac{E^2 R}{(R + r)^2} \]
Step 3: Condition for maximum power For maximum power, differentiate \(P\) with respect to \(R\) and equate to zero: \[ \frac{dP}{dR} = 0 \] \[ (R + r)^2 - 2R(R + r) = 0 \] \[ (R + r)(r - R) = 0 \] Since resistance cannot be negative, \[ R = r \] Final Answer: \[ \boxed{R = r} \]