An inductor of \(20~\text{mH}\), a capacitor of \(100~\mu \text{F}\), and a resistor of \(50~\Omega\) are connected in series across a source of emf, \(V=10 \sin (314 t)\). What is the power loss in this circuit?
1. \( 0.79 ~\text{W} \)
2. \( 0.43 ~\text{W} \)
3. \( 2.74 ~\text{W} \)
4. \( 1.13 ~\text{W}\)
A \(50\) Hz AC source of \(20\) volts is connected across \(R\) and \(C\) as shown in the figure below.
If the voltage across \(R\) is \(12\) volts, then the voltage across \(C\) will be:
1. | \(8\) V |
2. | \(16\) V |
3. | \(10\) V |
4. | not possible to determine unless values of \(R\) and \(C\) are given |
1. | \(10\) mH |
2. | \(100\) mH |
3. | \(1\) mH |
4. | Cannot be calculated unless \(R\) is known |
1. | AC cannot pass through DC Ammeter. |
2. | AC changes direction. |
3. | Average value of current for the complete cycle is zero. |
4. | DC Ammeter will get damaged. |
1. | circuit will be capacitive if \(\omega>\frac{1}{\sqrt{LC}} \) |
2. | circuit will be inductive if \(\omega=\frac{1}{\sqrt{LC}} \) |
3. | power factor of circuit will be unity if capacitive reactance equals inductive reactance |
4. | current will be leading voltage if \(\omega>\frac{1}{\sqrt{LC}} \) |
1. | \(\dfrac{E^2_0}{\sqrt{2}R}\) | 2. | \(\dfrac{E^2_0}{4R}\) |
3. | \(\dfrac{E^2_0}{2R}\) | 4. | \(\dfrac{E^2_0}{8R}\) |
A series AC circuit has a resistance of \(4~\Omega\) and an inductor of reactance \(3~\Omega\). The impedance of the circuit is \(z_1\). Now when a capacitor of reactance \(6~\Omega\) is connected in series with the above combination, the impedance becomes \(z_2\). Then \(\frac{z_1}{z_2}\) will be:
1. \(1:1\)
2. \(5:4\)
3. \(4:5\)
4. \(2:1\)
An inductor \((L)\) and resistance \((R)\) are connected in series with an AC source. The phase difference between voltage \((V)\) and current \((i)\) is \(45^{\circ}.\)
If the phase difference between \(V\) and \(i\) remains the same, then the capacitive reactance and impedance of the \(LCR\) circuit will be:
1. \(2R, R\sqrt{2}\)
2. \(R, R\sqrt{2}\)
3. \(R, R\)
4. \(2R, R\sqrt{3}\)
An \(AC\) voltage is applied to a resistance \(R\) and an inductor \(L\) in series. If \(R\) and the inductive reactance are both equal to \(3~ \Omega, \) then the phase difference between the applied voltage and the current in the circuit will be:
1. | \( \pi / 4\) | 2. | \( \pi / 2\) |
3. | zero | 4. | \( \pi / 6\) |
1. | Frequency of the AC source is decreased |
2. | The number of turns in the coil is reduced |
3. | A capacitance of reactance \(X_C = X_L\) is included in the same circuit |
4. | An iron rod is inserted in the coil |