Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged. Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor (R/√(R² + ω² L²)), where ω is frequency of the supply across resistor R and inductor L. If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage. In the light of the above statements, choose the most appropriate answer from the options given below: (a) (A) is true but (R) is false (b) Both (A) and (R) are true and (R) is the correct explanation of (A) 3. (c) Both (A) and (R) are true but (R) is not the correct explanation of (A) (d) (A) is false but (R) is true

Answer: (b). A choke coil has large L and small resistance, so it limits current with almost no power loss; it protects the tube from full mains voltage. (A) is true. With the choke in series, current I = (V)/(√(R² + ω²L²)), so tube voltage IR = V(R)/(√(R² + ω²L²)), reduced by that factor. Without the choke it would equal V. (R) is true. This voltage reduction is exactly why the tube is not damaged, so (R) explains (A). So the answer is (b).

Given below are two statements: one is labelled as Assertion (A) and… | JEE Main 2025 (29 Jan, Shift 1)

Q22JEE Main 2025 · 29 JAN · SHIFT 1Easy

Power, Power Factor & Wattless Current

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged.

Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor $\left(R/\sqrt{R^2 + \omega^2 L^2}\right)$ , where $\omega$ is frequency of the supply across resistor $R$ and inductor $L$ . If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage. In the light of the above statements, choose the most appropriate answer from the options given below:

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Answer: (b)

A choke coil has large $L$ and small resistance, so it limits current with almost no power loss; it protects the tube from full mains voltage. (A) is true.
With the choke in series, current $I = \frac{V}{\sqrt{R^2 + \omega^2L^2}}$ , so tube voltage $IR = V\frac{R}{\sqrt{R^2 + \omega^2L^2}}$ , reduced by that factor. Without the choke it would equal $V$ . (R) is true.
This voltage reduction is exactly why the tube is not damaged, so (R) explains (A).
So the answer is (b).

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