Q. No.Alternating current cannot be measured by a DC ammeter because:
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. | \(\dfrac{E^2_0}{\sqrt{2}R}\) | 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. \(2R, R\sqrt{3}\)
| 1. | \(10\) mH |
| 2. | \(100\) mH |
| 3. | \(1\) mH |
| 4. | Cannot be calculated unless \(R\) is known |
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. \(300\) V, \(15\) A
2. \(450\) V, \(15\) A
3. \(450\) V, \(13.5\) A
4. \(600\) V, \(15\) A
The core of a transformer is laminated because:
| 1. | Energy losses due to eddy currents may be minimized |
| 2. | The weight of the transformer may be reduced |
| 3. | Rusting of the core may be prevented |
| 4. | Ratio of voltage in primary and secondary may be increased |
| 1. | \(\frac{\varepsilon^{2} R}{\left[R^{2}+\left(L \omega-\frac{1}{C \omega}\right)^{2}\right]}\) | 2. | \(\frac{\varepsilon^{2} \sqrt{R^{2}+\left(L \omega-\frac{1}{C \omega}\right)^{2}}}{R}\) |
| 3. | \(\frac{\varepsilon^{2}\left[R^{2}+\left(L \omega-\frac{1}{C \omega}\right)^{2}\right]}{R}\) | 4. | \(\frac{\varepsilon^{2}R}{\sqrt{R^{2}+\left(L \omega-\frac{1}{C \omega}\right)^{2}}}\) |
A coil of inductive reactance of \(31~\Omega\) has a resistance of \(8~\Omega\). It is placed in series with a condenser of capacitive reactance \(25~\Omega\). The combination is connected to an AC source of \(110\) V. The power factor of the circuit is:
1. \(0.56\)
2. \(0.64\)
3. \(0.80\)
4. \(0.33\)
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