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 :

<p><strong>Assertion (A):</strong> </p> <blockquote> <p>A choke coil is simply a coil having large inductance but small resistance. </p> <p>Choke coils are used with fluorescent mercury-tube fittings. </p> <p>If household electric power is directly connected to a mercury tube, the tube will be damaged.</p> </blockquote> <p><p>A choke (or ballast) <strong>does</strong> have high inductance and low resistance. </p></p> <p><p>In a fluorescent lamp (or mercury‐vapor lamp) circuit, the choke limits the <strong>current</strong> through the tube. </p></p> <p><p>If we connect the tube directly to the full line voltage <strong>without</strong> any current-limiting device, excessive current will flow and damage the tube.</p> <p>Therefore, <strong>(A) is true</strong>.</p></p> <p><strong>Reason (R):</strong> </p> <blockquote> <p>By using the choke coil, the voltage across the tube is reduced by a factor </p> <p>$\displaystyle \frac{R}{\sqrt{R^2 + (\omega L)^2}}$, </p> <p>where $\omega$ is the supply frequency, and $R$ and $L$ are the resistance and inductance in series. </p> <p>If the choke coil were not used, the voltage across the resistor (tube) would be the same as the applied voltage.</p> </blockquote> <p><p>In a simple <strong>series R–L</strong> circuit (where $R$ represents the lamp’s effective resistance once conducting, and $L$ is the choke’s inductance), the total impedance is </p> <p>$ Z = \sqrt{\,R^2 + (\omega L)^2\,}. $ </p></p> <p><p>The current is the same through $R$ and $L$. The fraction of the <strong>total supply voltage</strong> appearing across $R$ alone is </p> <p>$ \frac{V_R}{V_{\text{total}}} \;=\; \frac{I\,R}{I\,Z} \;=\; \frac{R}{\sqrt{R^2 + (\omega L)^2}}. $ </p></p> <p><p>This shows that using a large inductance $L$ (the choke) can substantially reduce (or limit) the voltage—and hence the current—through the lamp. </p></p> <p><p>Without the choke coil, the tube (treated here as a resistor once it starts conducting) would indeed see nearly the <strong>full line voltage</strong>, causing a large current that could destroy it.</p> <p>Therefore, <strong>(R) is also true</strong>. </p></p> <p><strong>Does (R) correctly explain (A)?</strong> </p> <p><p>(A) states <em>why</em> a choke coil is needed (to prevent the lamp from being damaged by excessive current). </p></p> <p><p>(R) shows <em>how</em> the choke coil limits the voltage/current through the lamp—by giving the fraction of the supply voltage that appears across the lamp in a series R–L circuit. </p></p> <p><p>Thus, (R) <strong>does</strong> provide the correct (physics-based) explanation: the choke’s inductive reactance drops a significant portion of the supply voltage, thereby protecting the lamp.</p></p> <p>Hence, the correct option is:</p> <p>$ \boxed{\text{Option B: Both (A) and (R) are true and (R) is the correct explanation of (A).}} $</p>