An AC-voltage is applied across the ends \(A,B\) of the circuit shown in the figure, the RMS voltage being \(V_0\). It is observed that the RMS currents in all the three branches are equal to \(\dfrac{V_0}{R}\). The total current entering at \(A\) equals:
1. | \(\dfrac{V_0}{R}\) | 2. | \(\sqrt{2}\dfrac{V_0}{R}\) |
3. | \(\Big(\sqrt{2}+1\Big)\dfrac{V_0}{R}\) | 4. | \(\sqrt{3}\dfrac{V_0}{R}\) |
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The charge flowing through a resistance \(R\) is given by
\(q = A\sin\omega t+B\cos\omega t\)
The heat produced in the resistance over a very long time \(\Delta t\) (much greater than \(T = \dfrac{2\pi}{\omega}\)) is:
1. | \(\omega^{2}\left(A^{2}+B^{2}\right) R \Delta t\) |
2. | \(\omega^{2}\left(A^{2}-B^{2}\right) R \Delta t\) |
3. | \(\dfrac{1}{2} \omega^{2}\left(A^{2}+B^{2}\right) R \Delta t\) |
4. | \(\dfrac{1}{2} \omega^{2}\left(A^{2}-B^{2}\right) R \Delta t\) |
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An inductor \(200\) mH, capacitor \(20~\mu\text{F}\) and a resistance of \(200~\Omega\) are connected in series across a source of emf, \(E = (20~\text{V})\sin(500t)\), where \(t\) is in second. The power loss in the circuit is:
1. \(2~\text{W}\)
2. \(1~\text{W}\)
3. \(0.5~\text{W}\)
4. \(0.25~\text{W}\)
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1. | \(V_{0} I_{0} {\cos\phi}\) |
2. | \(\dfrac{V_{0} I_{0} \cos \phi}{2}\) |
3. | \(V_{0} I_{0} \sin \phi\) |
4. | \(\dfrac{V_{0} I_{0} \sin \phi}{2}\) |
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1. | zero | 2. | \(\sqrt 2 V_r \) |
3. | \(2 V_r\) | 4. | \(\dfrac{V_r}{\sqrt 2}\) |
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1. | \(\dfrac{V_r}{3}\) | 2. | \(\dfrac{2V_r}{3}\) |
3. | \(\dfrac{V_r}{2}\) | 4. | \(V_r\) |
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In the given scenario, the voltage, \(V_2 > V_1,\) and no current flow through the source on the left. The phase difference between the two sources is \(\phi.\)
1. | \(R\sin\phi= \dfrac{1}{\omega C}\) | 2. | \(R\cos\phi= \dfrac{1}{\omega C}\) |
3. | \(R\tan\phi= \dfrac{1}{\omega C}\) | 4. | \(R\cot\phi= \dfrac{1}{\omega C}\) |
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