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
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: