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1. Find the voltage across the resistor R if VA = -3V and VB = -5V. Use ideal diode model assumption. a) 0V
b) -3V
c) -5V
d) -4V

Answer: a [Reason:] In ideal diode model the diode is considered as a perfect conductor in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero. Since both diode is in reverse bias mode current through diode is zero.

2. Find current I if voltage V = 5V, VB = 2V, R1 & R2 = 2K. Use ideal diode model assumption. a) 1.5mA
b) 1.375mA
c) 2mA
d) 3mA

Answer: d [Reason:] In ideal diode model the diode is considered as a perfect conductor in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero. 4V forward biases the diode while 2V is reverse bias to the diode. So voltage across R1 is V-VB. That is 3V. Therefore, current through R1 = 3V/2k = 1.5mA. Current through R2 = 3/2K = 2.5mA. Therefore, total current I = 1.5+1.5 = 3mA.

3. Find current I if V = 5V and -5V when VB = 2V, R1 = 2K, R2 = 4K respectively are a) 0A and 1.3mA
b) 1.231mA and 0.33mA
c) 3.25mA and 0A
d) 1.58mA and 0A

Answer: c [Reason:] In ideal diode model the diode is considered as a perfect conductor in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero. V is forward biasing and VB is reverse biasing to diode. Current through resistor R1 = V/2k = 2.5mA. Current through resistor R2 = (V – VB ) /4k = 0.75mA. So total current is 3.25mA. At V = -5V, diode is reverse bias So the current is zero.

4. The output voltage V if Vin = 3V, R=5K, VB = 2V is a) 1V
b) 4V
c) 5V
d) 2.5V

Answer: a [Reason:] In ideal diode model the diode is considered as a perfect conductor in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero. Vin reverse biases the diode and VB so total voltage is Vin – VB.

5. In the circuit below VB = 2V, Vin = 5V. The voltage V across resistor R is a) 5V
b) 2V
c) 3V
d) 7V

Answer: b [Reason:] In ideal diode model the diode is considered as a perfect conductor in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero. Since Vin reverse biases the diode it won’t be shown in resistor R. So only VB appears.

6. In the circuit Vin = 4V, VB = 3V, R = 5K. The voltage across diode V is a) 1V
b) 4V
c) 3V
d) 7V

Answer: a [Reason:] In ideal diode model the diode is considered as a perfect conductor in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero. Since Vin reverse biases the diode and VB forward biases the diode So total voltage across diode is Vin-VB.

7. In the circuit below Vin = 4V, R = 2K and VB = 2V. In these conditions the voltage across diode V is a) -4V
b) -2V
c) 2V
d) 0V

Answer: d [Reason:] In ideal diode model the diode is considered as a perfect conductor in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero. Since net voltage Vin – VB = 2V forward biases the diode voltage across diode is zero.

8. In the circuit shown in below I = 2mA, VB = 2V and R = 2K. The voltage V will be a) 0V
b) 2V
c) 4V
d) 1V

Answer: b [Reason:] In ideal diode model the diode is considered as a perfect conductor in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero. Since current I is forward bias to diode resistor R has no relevance so output voltage is VB.

9. In the circuit shown in below I = 2mA, VB = 2V and R = 2K. The voltage V will be a) 2V
b) 3V
c) 6V
d) 5V 