1. Voltage across the resistor R if V_{A} = -3V and V_{B} = -5V is

(Use constant voltage drop model assumption and take V_{D} = 0.5V)

a) 0V

b) -3V

c) -5V

d) -4V

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2. Current I if voltage V = 5V, V_{B} = 2V, R_{1} & R_{2} = 2K

(Use constant voltage drop model assumption and take V_{D} = 0.5V)

a) 1.25mA

b) 1mA

c) 2.75mA

d) 1.75mA

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_{in}and V

_{B}are opposite net voltage is 3V. Voltage at R

_{1}is 3V so current is 1.5mA. Voltage at R

_{2}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 V_{B} = 2V, R_{1} = 2K, R_{2} = 4K respectively are

(Use constant voltage drop model assumption and take V_{D} = 0.5V)

a) 1.3mA, 0.23mA

b) 2.875mA, 0mA

c) 2mA, 0mA

d) 1.423mA, 0 mA

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4. The output voltage V if V_{in} = 3V, R=5K, V_{B} = 2V is

(Use constant voltage drop model assumption and take V_{D} = 0.5V)

a) 0V

b) 3.5V

c) 2.5V

d) 1.5V

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_{in}is reverse bias to the diode all V

_{in}will appear across diode. Since V

_{in}is opposite to V

_{B}net voltage will be V

_{in}-(V

_{B}-V

_{D}).

5. In the circuit below V_{B} = 2V, V_{in} = 5V. The voltage V across resistor R is

(Use constant voltage drop model assumption and take V_{D} = 0.5V)

a) 1.5V

b) 2.5V

c) 3.5V

d) 2V

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_{in}reverse bias to the diode it has no effect in R. Therefore, V

_{B}alone control voltage at R. So voltage across R will be same as V

_{B}with a voltage drop of V

_{D}.

6. In the circuit V_{in} = 4V, V_{B} = 3V, R = 5K. The voltage across diode V is

(Use constant voltage drop model assumption and take V_{D} = 0.5V)

a) 0V

b) 0.5V

c) 1V

d) 1.5V

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7. In the circuit below V_{in} = 4V, R = 2K and V_{B} = 2V. In these conditions the voltage across diode V is

(Use constant voltage drop model assumption and take V_{D} = 0.5V)

a) 2.5

b) 4.5

c) 0.5

d) 1.5

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_{D}.

8. In the circuit shown in below I = 2mA, V_{B} = 2V and R = 2K. The voltage V will be

(Use constant voltage drop model assumption and take V_{D} = 0.5V)

a) 0V

b) 2.5V

c) 6V

d) 3.5V

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_{D}. So net voltage output is V

_{D}+V

_{B}.

9. In the circuit shown in below I = 2mA, V_{B} = 2V and R = 2K. The voltage V will be

(Use constant voltage drop model assumption and take V_{D} = 0.5V)

a) 4V

b) 3V

c) 6V

d) 5.5V

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10. For circuit shown below V_{in} = 3V, R_{1} = 6K, R_{2} = 2K. The voltage V will be

(Use constant voltage drop model assumption and take V_{D} = 0.5V)

a) 2V

b) 3V

c) 3.5V

d) 2.5V

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_{1}voltage drop across diode is V

_{D}. So net voltage equals to V

_{in}–V

_{D}.