- 1(current)
Q. No.The current through the inductor in the figure is initially zero. The initial rate of change of the current \(i\) through the inductor (i.e. \(\dfrac{di}{dt}\)) is:
| 1. | zero | 2. | \(-\dfrac{I_{0} R}{L}\) |
| 3. | \(\dfrac{I_{0} R}{L}\) | 4. | \(\dfrac{I_{0} R}{2L}\) |
Subtopic: Â LR circuit |
From NCERT
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A Blackbox (BB) which may contain a combination of electrical circuit elements (resistor, capacitor or inductor) is connected with other external circuit elements as shown below in figure (a). After the switch is closed at time \(t=0,\), the current \((I)\) as a function of time \((t)\) is shown in figure (b).
From this, we can infer that the Blackbox contains:
| 1. | a resistor and a capacitor in series. |
| 2. | a resistor and a capacitor in parallel. |
| 3. | a resistor and an inductor in series. |
| 4. | a resistor and an inductor in parallel. |
Subtopic: Â LR circuit |
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From NCERT
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An inductor of \(0.1\) H is connected with a \(10~\Omega\) resistance in series with it, as shown in the figure. A dc-source of emf \(20\) V and an ac-source of emf \(20\) V (rms) are separately connected across \(AB.\) The ac-frequency is \(\frac{50}{\pi}\)Hz.
Let the dc-current be \(i\) (after a long time) and the peak ac-current be \(i_p.\) The ratio \(\frac{i}{i_p}\) equals
1. \(1\)
2. \(\sqrt2\)
3. \(\frac{1}{\sqrt2}\)
4. \(\frac12\)
Let the dc-current be \(i\) (after a long time) and the peak ac-current be \(i_p.\) The ratio \(\frac{i}{i_p}\) equals
1. \(1\)
2. \(\sqrt2\)
3. \(\frac{1}{\sqrt2}\)
4. \(\frac12\)
Subtopic: Â LR circuit |
From NCERT
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An inductor of \(0.1~\text H\) is connected with a \(10~\Omega\) resistance in series with it, as shown in the figure. A DC-source of emf \(20~\text V\) and an AC-source of emf \(20~\text{V}\) (RMS) are separately connected across \(AB.\) The ac-frequency is \(\dfrac{50}{\pi}~\text{Hz}.\)
Let the DC-energy stored in the inductor be \(U\) (after a long time) and the average energy stored in the inductor (with the ac-source) be \(U_{av}.\) The ratio \(\dfrac{U_{av}}{U}\) equals:
1. zero
2. \(1\)
3. \(\dfrac12\)
4. \(\dfrac14\)
Let the DC-energy stored in the inductor be \(U\) (after a long time) and the average energy stored in the inductor (with the ac-source) be \(U_{av}.\) The ratio \(\dfrac{U_{av}}{U}\) equals:
1. zero
2. \(1\)
3. \(\dfrac12\)
4. \(\dfrac14\)
Subtopic: Â LR circuit |
From NCERT
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- 1(current)
Q. No.
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