At resonance the phase angle equals

Here are the solutions to the multiple-choice questions:

9. At resonance the phase angle equals

• At resonance in an RLC circuit, the inductive reactance ($X_L$) cancels out the capacitive reactance ($X_C$). The circuit behaves purely resistively.
• Therefore, the voltage and current are in phase, and the phase angle ($\phi$) is $0^\circ$.
• Answer: a) $\boxed{\text{0 Degree}}$

10. The band width of resonant circuit is

• The bandwidth ($\Delta f$) of a resonant circuit is inversely proportional to its quality factor ($Q$). The formula is $\Delta f = \frac{f_r}{Q}$, where $f_r$ is the resonant frequency.
• Answer: b) $\boxed{\text{Inversely proportional to Q}}$

11. In silicon atom distribution of electrons in cells is

• Silicon (Si) has an atomic number of 14. Its electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^2$.
• The electron distribution in shells (K, L, M) is:
  • K-shell (1st shell): 2 electrons
  • L-shell (2nd shell): $2+6 = 8$ electrons
  • M-shell (3rd shell): $2+2 = 4$ electrons
• So the distribution is 2, 8, 4.
• Answer: a) $\boxed{\text{2,8,4}}$

12. An inductive circuit with $L=100 \text{mH}$ and $R=10 \text{K}\Omega$ has time constant of

• The time constant ($\tau$) for an RL circuit is given by the formula $\tau = \frac{L}{R}$.
• Given $L = 100 \text{ mH} = 100 \times 10^{-3} \text{ H} = 0.1 \text{ H}$.
• Given $R = 10 \text{ K}\Omega = 10 \times 10^3 \text{ } \Omega$.
• $\tau = \frac{0.1 \text{ H}}{10 \times 10^3 \text{ } \Omega} = \frac{0.1}{10000} \text{ s} = 0.00001 \text{ s}$
• $\tau = 10 \times 10^{-6} \text{ s} = 10 \text{ µs}$
• Answer: d) $\boxed{\text{10 µsec}}$

13. A capacitive circuit with $R=15 \text{K}\Omega$ and $C=0.01 \text{µF}$ has a time constant of

• The time constant ($\tau$) for an RC circuit is given by the formula $\tau = RC$.
• Given $R = 15 \text{ K}\Omega = 15 \times 10^3 \text{ } \Omega$.
• Given $C = 0.01 \text{ µF} = 0.01 \times 10^{-6} \text{ F} = 1 \times 10^{-8} \text{ F}$.
• $\tau = (15 \times 10^3 \text{ } \Omega) \times (1 \times 10^{-8} \text{ F})$
• $\tau = 15 \times 10^{-5} \text{ s} = 0.00015 \text{ s}$
• $\tau = 150 \times 10^{-6} \text{ s} = 150 \text{ µs}$
• Answer: c) $\boxed{\text{150 µsec}}$

14. The development of voltage difference due to light intensity on the surface of a PN junction is known

• When light falls on a PN junction, it generates electron-hole pairs, leading to a voltage difference across the junction. This phenomenon is called the photovoltaic effect and is the basis for solar cells.
• Answer: b) $\boxed{\text{Photo voltaic effect}}$

15. The function of choke in fluorescent tube is

• A choke (inductor) in a fluorescent tube circuit performs two main functions:
  • It provides a high voltage spike (strike voltage) to ionize the gas and start the discharge when the circuit is switched on.
  • Once the tube is lit, it limits the current flowing through the tube to a safe operating level.
• Answer: c) $\boxed{\text{Both a and b are correct}}$