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 \( \frac{R}{\sqrt{R^2 + \omega^2 L^2}} \), where \( \omega \) is the 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 light of the above statements, choose the most appropriate answer from the options given below:
Assertion (A): 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 \( \frac{R}{\sqrt{R^2 + \omega^2 L^2}} \), where \( \omega \) is the 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.
Explanation:
The assertion (A) is true. A choke coil indeed has a large inductance and small resistance. It is used in fluorescent tube lights primarily to limit the current in the circuit without a significant amount of power loss, which typical resistors would cause. Connecting a mercury tube directly to household power can cause excessive current and damage the tube.
However, the reason (R) is false. The formula given \( \frac{R}{\sqrt{R^2 + \omega^2 L^2}} \) pertains to the current division principle in AC circuits involving reactance and resistance. The choke coil does not primarily function by reducing the voltage in this manner; instead, it mainly controls the current using its inductive reactance which depends on the frequency \(\omega\) and the inductance \(L\). The voltage drop across an inductive choke is not solely responsible for protecting the fluorescent tube; rather, the inductive reactance limits the current flow, preventing damage.
Thus, the most appropriate answer is:
(A) is true but (R) is false
An air filled parallel plate electrostatic actuator is shown in the figure. The area of each capacitor plate is $100 \mu m \times 100 \mu m$. The distance between the plates $d_0 = 1 \mu m$ when both the capacitor charge and spring restoring force are zero as shown in Figure (a). A linear spring of constant $k = 0.01 N/m$ is connected to the movable plate. When charge is supplied to the capacitor using a current source, the top plate moves as shown in Figure (b). The magnitude of minimum charge (Q) required to momentarily close the gap between the plates is ________ $\times 10^{-14} C$ (rounded off to two decimal places). Note: Assume a full range of motion is possible for the top plate and there is no fringe capacitance. The permittivity of free space is $\epsilon_0 = 8.85 \times 10^{-12} F/m$ and relative permittivity of air ($\epsilon_r$) is 1.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 