Q. No.The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance \(R_1\) will be:
1. \(2.5~\text{A}\)
2. \(10.0~\text{A}\)
3. \(1.43~\text{A}\)
4. \(3.13~\text{A}\)
1. \(2.5~\text{A}\)
2. \(10.0~\text{A}\)
3. \(1.43~\text{A}\)
4. \(3.13~\text{A}\)
| 1. | Electrons are minority carriers and pentavalent atoms are dopants. |
| 2. | Holes are minority carriers and pentavalent atoms are dopants. |
| 3. | Holes are the majority carriers and trivalent atoms are dopants. |
| 4. | Electrons are the majority carriers and trivalent atoms are dopants. |
The output \((X)\) of the logic circuit shown in the figure will be:
1. \(X= \overline{A\cdot B}\)
2. \(X = A\cdot B\)
3. \(X= \overline{A+ B}\)
4. None of the above
Transfer characteristics [output voltage () vs input voltage ()] for a base biased transistor in CE configurations are as shown in the figure. For using the transistor as a switch, it is used:
1. In region III
2. Both in the region (I) and (III)
3. In region II
4. In region I
| 1. | the positive terminal of the battery is connected to the p-side and the depletion region becomes thick. |
| 2. | the negative terminal of the battery is connected to the n-side and the depletion region becomes thin. |
| 3. | the positive terminal of the battery is connected to the n-side and the depletion region becomes thin. |
| 4. | the negative terminal of the battery is connected to the p-side and the depletion region becomes thick. |
To get an output Y = 1 from the circuit shown below, the input must be:
1. A=0 B=1 C=0
2. A=0 B=0 C=1
3. A=1 B=0 C=1
4. A=1 B=0 C=0
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