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Q. No.In half-wave rectification, if the input frequency is \(60\) Hz, the output frequency will be:
1.
\(120\) Hz
2.
zero
3.
\(30\) Hz
4.
\(60\) Hz
Subtopic: Â Rectifier |
 67%
Level 2: 60%+
NEET - 2022
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As the temperature increases, the electrical resistance:
| 1. | decreases for conductors but increases for semiconductors. |
| 2. | increases for both conductors and semiconductors. |
| 3. | decreases for both conductors and semiconductors. |
| 4. | increases for conductors but decreases for semiconductors. |
Subtopic: Â Energy Band theory |
 75%
Level 2: 60%+
NEET - 2022
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In the given circuits \((a),\) \((b),\) and \((c),\) the potential drop across the two \({\mathrm{p\text-n}}\) junctions is equal in:
| 1. | both circuits \((a)\) and \((c)\) |
| 2. | circuit \((a)\) only |
| 3. | circuit \((b)\) only |
| 4. | circuit \((c)\) only |
Subtopic: Â PN junction |
 65%
Level 2: 60%+
NEET - 2022
Hints
In the circuit shown in the figure, \(V_A>V_B.\) Which of the diodes conduct?
| 1. | only \(D_3\) | 2. | \(D_1\) and \(D_3\) |
| 3. | all of the diodes | 4. | none of the diodes |
Subtopic: Â PN junction |
 53%
Level 3: 35%-60%
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When a \(\mathrm{p\text-}\)type semi-conductor is put in an electric field \(\vec{E}\) the electrons in the valence band, on average:
| 1. | flow in the direction of the \(\vec{E}\) and cause a current opposite to \(\vec{E}\) |
| 2. | flow opposite to \(\vec{E}\) and cause a current along \(\vec{E}\) |
| 3. | flow along \(\vec{E}\) and cause a current along \(\vec{E}\) |
| 4. | flow opposite to \(\vec{E}\) and cause a current opposite to \(\vec{E}\) |
Subtopic: Â Types of Semiconductors |
 64%
Level 2: 60%+
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The resistance between \(A\) and \(B\) is
| 1. | \(100~\Omega\) if \(V_A>V_B\) and \(200~\Omega\) if \(V_A<V_B\) |
| 2. | \(100~\Omega\) if \(V_A<V_B\) and \(200~\Omega\) if \(V_A>V_B\) |
| 3. | \(100~\Omega\) |
| 4. | \(200~\Omega\) |
Subtopic: Â PN junction |
 74%
Level 2: 60%+
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The circuit shown in figure is given an input signal \(V_i,\) which varies with time and the corresponding output is \(V_o.\) Then,
| 1. | \(V_o=V_i+E\) |
| 2. | \(V_o=V_i-E\) |
| 3. | \(V_o=V_i,\) only when \(V_i>E\) |
| 4. | \(V_o=V_i,\) only when \(V_i<E\) |
Subtopic: Â PN junction |
Level 3: 35%-60%
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The voltage \(V_{AB}=20~\text{V}\) at its peak value, and is sinusoidal in time. The current \(i\) (in amperes), when plotted as a function of time, is given by:
(consider the diodes as ideal.)
(consider the diodes as ideal.)
| 1. |  |
2. |  |
| 3. |  |
4. |  |
Subtopic: Â PN junction |
Level 3: 35%-60%
Hints
The current-voltage characteristic of an ideal \(\mathrm{p \text{-}n}\) junction diode is given by the graph as shown in the following figure:
This diode is connected with a resistance of \(5~\Omega\) in series with it as shown below:
Which of the following shows the dependence of the voltage \(V_{AB}\) and the current \(i,\) when the diode is forward-biased?
(\(V_{AB}\) in volt, \(i\) in ampere)
This diode is connected with a resistance of \(5~\Omega\) in series with it as shown below:
Which of the following shows the dependence of the voltage \(V_{AB}\) and the current \(i,\) when the diode is forward-biased?
(\(V_{AB}\) in volt, \(i\) in ampere)
| 1. | \(V_{A B}=i\cdot5+0.6\) |
| 2. | \(V_{A B}=i\cdot5-0.6\) |
| 3. | \(V_{A B}=i\cdot5+(0.6-5)\) |
| 4. | \(V_{A B}=i\cdot5+\left(0.6+5\right)\) |
Subtopic: Â PN junction |
Level 3: 35%-60%
Hints
A \(\mathrm{p\text-n}\) junction has an electric field of \(6\times 10^{5}~\text{V/m}\) in the junction and the junction width is \(500~\text{nm}.\) The height of the potential barrier is:
1. \(0.6~\text V\)
2. \(0.3~\text V\)
3. \(0.5~\text V\)
4. \(0.25~\text V\)
1. \(0.6~\text V\)
2. \(0.3~\text V\)
3. \(0.5~\text V\)
4. \(0.25~\text V\)
Subtopic: Â PN junction |
 88%
Level 1: 80%+
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