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1. Voltage across the resistor R if VA = -3V and VB = -5V is
(Use constant voltage drop model assumption and take VD = 0.5V)
analog-circuits-questions-answers-experienced-q1
a) 0V
b) -3V
c) -5V
d) -4V

View Answer

Answer: a [Reason:] In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open.

2. Current I if voltage V = 5V, VB = 2V, R1 & R2 = 2K
(Use constant voltage drop model assumption and take VD = 0.5V)
analog-circuits-questions-answers-experienced-q2
a) 1.25mA
b) 1mA
c) 2.75mA
d) 1.75mA

View Answer

Answer: c [Reason:] In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open. Since Vin and VB are opposite net voltage is 3V. Voltage at R1 is 3V so current is 1.5mA. Voltage at R2 is 3-0.5 = 2.5V. So the current is 1.25mA. The net current is 2.75mA.

3. Current I if V = 5V and -5V when VB = 2V, R1 = 2K, R2 = 4K respectively are
(Use constant voltage drop model assumption and take VD = 0.5V)
analog-circuits-questions-answers-experienced-q3
a) 1.3mA, 0.23mA
b) 2.875mA, 0mA
c) 2mA, 0mA
d) 1.423mA, 0 mA

View Answer

Answer: b [Reason:] In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open. Since diode is in forward bias net total voltage becomes 4.5V. Current through branch 1 will be 4.5V / 2K = 2.25mA. Current through branch 2 will be (4.5-2)/4K = 0.625mA. So net current is sum of these two. Therefore, net current is 2.875mA.

4. The output voltage V if Vin = 3V, R=5K, VB = 2V is
(Use constant voltage drop model assumption and take VD = 0.5V)
analog-circuits-questions-answers-experienced-q4
a) 0V
b) 3.5V
c) 2.5V
d) 1.5V

View Answer

Answer: d [Reason:] In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open. Since Vin is reverse bias to the diode all Vin will appear across diode. Since Vin is opposite to VB net voltage will be Vin-(VB-VD).

5. In the circuit below VB = 2V, Vin = 5V. The voltage V across resistor R is
(Use constant voltage drop model assumption and take VD = 0.5V)
analog-circuits-questions-answers-experienced-q5
a) 1.5V
b) 2.5V
c) 3.5V
d) 2V

View Answer

Answer: a [Reason:] In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open. Since Vin reverse bias to the diode it has no effect in R. Therefore, VB alone control voltage at R. So voltage across R will be same as VB with a voltage drop of VD.

6. In the circuit Vin = 4V, VB = 3V, R = 5K. The voltage across diode V is
(Use constant voltage drop model assumption and take VD = 0.5V)
analog-circuits-questions-answers-experienced-q6
a) 0V
b) 0.5V
c) 1V
d) 1.5V

View Answer

Answer: c [Reason:] In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open. Since diode is in reverse bias mode, the net voltage will appear on the voltage that is 1V.

7. In the circuit below Vin = 4V, R = 2K and VB = 2V. In these conditions the voltage across diode V is
(Use constant voltage drop model assumption and take VD = 0.5V)
analog-circuits-questions-answers-experienced-q7
a) 2.5
b) 4.5
c) 0.5
d) 1.5

View Answer

Answer: c [Reason:] In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open. Since diode is forward biased it will produce a voltage drop of VD.

8. In the circuit shown in below I = 2mA, VB = 2V and R = 2K. The voltage V will be
(Use constant voltage drop model assumption and take VD = 0.5V)
analog-circuits-questions-answers-experienced-q8
a) 0V
b) 2.5V
c) 6V
d) 3.5V

View Answer

Answer: b [Reason:] In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open. Since current source forward bases the diode voltage drop across diode is VD. So net voltage output is VD+VB.

9. In the circuit shown in below I = 2mA, VB = 2V and R = 2K. The voltage V will be
(Use constant voltage drop model assumption and take VD = 0.5V)
analog-circuits-questions-answers-experienced-q9
a) 4V
b) 3V
c) 6V
d) 5.5V

View Answer

Answer: c [Reason:] In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open. Since current source reverse biases diode, all current pass through resistor R. So voltage across resistor is 4V. Since voltage source cannot produce current due to lack of closed circuit total voltage at the output is 6V.

10. For circuit shown below Vin = 3V, R1 = 6K, R2 = 2K. The voltage V will be
(Use constant voltage drop model assumption and take VD = 0.5V)
analog-circuits-questions-answers-experienced-q10
a) 2V
b) 3V
c) 3.5V
d) 2.5V

View Answer

Answer: d [Reason:] In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open. Since diode is forward biased and parallel to resistor R1 voltage drop across diode is VD. So net voltage equals to Vin –VD.

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