A storage battery of emf 12\ V and internal resistance 0.5\ Ω is conn…

The correct answer is Option (3) → (A), (B) and (C) only

Note: The given answer is as per the official NTA Answer Key. However, none of the given options appears to be correct, because statement (C) is true only for the current at maximum power transfer, which the question does not state. Strictly, only statements (A) and (B) are correct, and this combination is not among the options, as explained below.

(A) Maximum power that can be delivered to the resistance \( R \) is \( 72\ \mathrm{W} \) — Correct. By the maximum power transfer theorem, the power in \( R \) is greatest when \( R = r \).

    \( \displaystyle P_{\max} = \frac{E^{2}}{4r} = \frac{(12\ \mathrm{V})^{2}}{4 \times 0.5\ \Omega} = 72\ \mathrm{W} \)

(B) Maximum current that can be drawn from the battery is \( 24\ \mathrm{A} \) — Correct. The current is largest when the external resistance is zero (short circuit).

    \( \displaystyle I_{\max} = \frac{E}{r} = \frac{12\ \mathrm{V}}{0.5\ \Omega} = 24\ \mathrm{A} \)

(C) Maximum current that can be drawn through the resistance \( R \) is \( 12\ \mathrm{A} \). — Correct as per NTA; however, it is not correct as a general statement. The whole circuit current flows through \( R \), so its largest value is also \( 24\ \mathrm{A} \) (as \( R \to 0 \)); 12 A is the current at maximum power, \( R = r \):

    \( \displaystyle I = \frac{12\ \mathrm{V}}{0.5\ \Omega + 0.5\ \Omega} = 12\ \mathrm{A} \)

(D) Maximum current that can be drawn through the resistance \( R \) is \( 6\ \mathrm{A} \) — Incorrect. 6 A flows only when \( R = 1.5\ \Omega \); it is neither the largest current nor the current at maximum power.

It seems that NTA has read statement (C) as the current through \( R \) when maximum power is delivered to it.