Question

A battery with EMF $E$ and internal resistance $r$ is connected across a resistance $R$. The power consumption in $R$ will be maximum when :

• A. $R = r$
• B. $R = r/2$
• C. $R = \sqrt{2} r$
• D. $R = 2r$

Hint:

Maximum Power Transfer Theorem: Power is max when $R_{ext} = R_{int}$. The efficiency at this point is exactly 50%.

Answer 1

The Correct Option is A

To determine the condition under which the power consumed by a resistor \(R\) connected to a battery of EMF \(E\) and internal resistance \(r\) is maximized, we use the Maximum Power Transfer Theorem.

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Step 1: Expression for Power Delivered to the Resistor

The current in the circuit is:

\[ I = \frac{E}{R + r} \]

The power consumed by the resistor \(R\) is:

\[ P = I^2 R = \frac{E^2 R}{(R + r)^2} \]

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Step 2: Condition for Maximum Power

To find the maximum power, differentiate \(P\) with respect to \(R\) and equate to zero:

\[ \frac{d}{dR}\left(\frac{E^2 R}{(R + r)^2}\right) = 0 \]

Solving this condition gives:

\[ R = r \]

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Step 3: Conclusion

Hence, the power delivered to the resistor is maximum when the external resistance equals the internal resistance of the battery.

\(\boxed{R = r}\)

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Step 4: Verification of Options

• \(R = r\) ✓ (Correct)
• \(R = \frac{r}{2}\) ✗
• \(R = \sqrt{2}\,r\) ✗
• \(R = 2r\) ✗