Solve the following multiple-choice questions: 1. A frequency of 2 MHz has time period of? 2. The 120 Volts r.m.s. has peak value of? 3. If the capacitance is doubled the capacitive reactance is? 4. T

Here are the solutions to the multiple-choice questions:

1. A frequency of $2 \text{ MHz}$ has time period of

• The relationship between frequency ($f$) and time period ($T$) is $T = \frac{1}{f}$.
• Given $f = 2 \text{ MHz} = 2 \times 10^6 \text{ Hz}$.
• $T = \frac{1}{2 \times 10^6 \text{ Hz}} = 0.5 \times 10^{-6} \text{ s} = 0.5 \text{ µs}$
• Answer: a) $\boxed{\text{0.5 µsec}}$

2. The $120 \text{ Volts r.m.s.}$ has peak value of

• For a sine wave, the peak voltage ($V_{peak}$) is related to the RMS voltage ($V_{rms}$) by $V_{peak} = V_{rms} \times \sqrt{2}$.
• Given $V_{rms} = 120 \text{ V}$.
• $V_{peak} = 120 \text{ V} \times \sqrt{2} \approx 120 \text{ V} \times 1.414 \approx 169.68 \text{ V}$
• This is approximately $170 \text{ V}$.
• Answer: b) $\boxed{\text{170 Volts}}$

3. If the capacitance is doubled the capacitive reactance is

• Capacitive reactance ($X_C$) is given by the formula $X_C = \frac{1}{2\pi f C}$.
• If capacitance $C$ is doubled to $2C$, the new capacitive reactance $X_C'$ will be:
• $X_C' = \frac{1}{2\pi f (2C)} = \frac{1}{2} \left(\frac{1}{2\pi f C}\right) = \frac{1}{2} X_C$
• Therefore, the capacitive reactance is halved.
• Answer: a) $\boxed{\text{Halved}}$

4. Two $5000 \text{ } \Omega$ inductive reactance are in parallel have an equivalent reactance of

• For two parallel reactances $X_{L1}$ and $X_{L2}$, the equivalent reactance $X_{eq}$ is given by:
• $\frac{1}{X_{eq}} = \frac{1}{X_{L1}} + \frac{1}{X_{L2}}$
• Given $X_{L1} = 5000 \text{ } \Omega$ and $X_{L2} = 5000 \text{ } \Omega$.
• $\frac{1}{X_{eq}} = \frac{1}{5000 \text{ } \Omega} + \frac{1}{5000 \text{ } \Omega} = \frac{2}{5000 \text{ } \Omega}$
• $X_{eq} = \frac{5000 \text{ } \Omega}{2} = 2500 \text{ } \Omega$
• Answer: a) $\boxed{\text{2500 } \Omega}$

5. In a sine wave ac circuit the voltage across a capacitance

• In a purely capacitive AC circuit, the current leads the voltage by $90^\circ$. Conversely, the voltage across the capacitance lags the current by $90^\circ$.
• Answer: a) $\boxed{\text{Lags current by } 90^\circ}$

6. Alternating current can induce voltage because alternating current has a

• According to Faraday's Law of Induction, a voltage is induced when there is a change in magnetic flux. Alternating current continuously changes its magnitude and direction, which produces a continuously varying magnetic flux.
• Answer: b) $\boxed{\text{Varying magnetic flux}}$

7. An iron core transformer with a $1:8$ step up ratio has $120 \text{V}$ applied across the primary. The voltage across the secondary equals

• For an ideal transformer, the ratio of secondary voltage ($V_s$) to primary voltage ($V_p$) is equal to the ratio of secondary turns ($N_s$) to primary turns ($N_p$):
• $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
• Given turns ratio $N_p : N_s = 1:8$, so $\frac{N_s}{N_p} = \frac{8}{1}$.
• Given primary voltage $V_p = 120 \text{ V}$.
• $\frac{V_s}{120 \text{ V}} = \frac{8}{1}$
• $V_s = 120 \text{ V} \times 8 = 960 \text{ V}$
• Answer: d) $\boxed{\text{960V}}$

8. When capacitance is increased the resonant frequency of the LC circuit

• The resonant frequency ($f_r$) of an LC circuit is given by the formula:
• $f_r = \frac{1}{2\pi\sqrt{LC}}$
• From the formula, $f_r$ is inversely proportional to the square root of the capacitance ($C$). If $C$ increases, $f_r$ will decrease.
• Answer: b) $\boxed{\text{Decreases}}$