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Semiconductor Electronics Material Devices and Simple Circuits

150749 For the given circuit shown below, to act as full wave rectifier, the $AC$ input should be connected across and and the DC. output would appear across and
original image

1 B and D and A and C

2 B and A and C and D

3 C and A and B and D

4 C and D and B and A

5 none of the above

Explanation:

B Let naming the diodes $D_{1}, D_{2}, D_{3}$ and $D_{4}$
original image
According to the question circuit is act as full wave rectifier.
In above figure, when we apply $AC$ input across $BD$ . For positive half cycle diode $D_{1} & D_{3}$ act as forward bias and $D_{2} & D_{4}$ act as reverse bias.
For negative half cycle $D_{1} & D_{3}$ act as a reverse bias and $D_{2} & D_{4}$ act as forward bias.
Hence, AC input should be connected across B and D and the DC. Output would appear across A and C.

Kerala CEE 2004

Semiconductor Electronics Material Devices and Simple Circuits

150750 In the circuit of figure, treat diode as ideal, current in the $4 Ω$ resistor is

1

2 A

2

3 A

3

\frac{12}{7} A

4

\frac{30}{13} A

Explanation:

B original image
From above figure, we can see diode $D_{1}$ is forward bias so, current flow through it and $D_{2}$ is reverse bias and it acts like a open circuit.
Then resistance $2 Ω$ and $4 Ω$ are in series connection.
$∴ R_{eq} = 2 + 4 = 6 Ω$
Now, current in $4 Ω$ resistance is
$I = \frac{V}{R_{e q}}$
$I = \frac{12}{6} = 2 A$

JCECE-2017]![original image](https://images.teachhire.in/YCT%20PHYSICS/14.%20SEMICONDUCTOR%20ELECTRONICS-%20MATERIALS-%20DEVICES%20AND%20SIMPLE%20CIRCUITS/CLFBqnRIYnI4klcFUltGCy-E1jk4q0hhda23jWA6G7Y.original.fullsize.png)

Semiconductor Electronics Material Devices and Simple Circuits

150752 A diode is connected to $220 V$ (rms) $AC$ in series with a capacitor as shown in figure. The voltage across the capacitor is
original image

1

220 V

2

110 V

3

311.1 V

4

\frac{220}{\sqrt{2}} V

Explanation:

D In this circuit diode D conduct in positive half cycle and in negative half cycle this is open circuit so it behaves as half wave rectifier.
The voltage across the capacitor,
$V_{rms} = \frac{V_{0}}{2} = \frac{220 \sqrt{2}}{2} \Rightarrow \frac{220}{\sqrt{2}}$

JCECE-2012

Semiconductor Electronics Material Devices and Simple Circuits

150761 In a transistor connected in a common emitter $mode, R_{C} = 4 k Ω, R_{1} = 1 k Ω, I_{C} = 1 mA$ and $I_{b} =$ $20 μ A$ . The voltage gain is

1 100

2 200

3 300

4 400

Explanation:

B Given that,
$R_{c} = 4 \times 10^{3} Ω R_{1} = 1 \times 10^{3} Ω$
$I_{C} = 1 \times 10^{- 3} I_{B} = 20 \times 10^{- 6} A$
We know that, voltage gain,
$A_{V} = \frac{β R_{C}}{R_{1}} = \frac{I_{C}}{I_{B}} \frac{R_{c}}{R_{1}} (∵ β = \frac{I_{C}}{I_{B}})$
$A_{v} = \frac{1 \times 10^{- 3} \times 4 \times 10^{3}}{20 \times 10^{- 6} \times 1 \times 10^{3}}$
$A_{V} = 200$

COMEDK 2015

Semiconductor Electronics Material Devices and Simple Circuits

150749 For the given circuit shown below, to act as full wave rectifier, the $AC$ input should be connected across and and the DC. output would appear across and
original image

1 B and D and A and C

2 B and A and C and D

3 C and A and B and D

4 C and D and B and A

5 none of the above

Explanation:

B Let naming the diodes $D_{1}, D_{2}, D_{3}$ and $D_{4}$
original image
According to the question circuit is act as full wave rectifier.
In above figure, when we apply $AC$ input across $BD$ . For positive half cycle diode $D_{1} & D_{3}$ act as forward bias and $D_{2} & D_{4}$ act as reverse bias.
For negative half cycle $D_{1} & D_{3}$ act as a reverse bias and $D_{2} & D_{4}$ act as forward bias.
Hence, AC input should be connected across B and D and the DC. Output would appear across A and C.

Kerala CEE 2004

Semiconductor Electronics Material Devices and Simple Circuits

150750 In the circuit of figure, treat diode as ideal, current in the $4 Ω$ resistor is

1

2 A

2

3 A

3

\frac{12}{7} A

4

\frac{30}{13} A

Explanation:

B original image
From above figure, we can see diode $D_{1}$ is forward bias so, current flow through it and $D_{2}$ is reverse bias and it acts like a open circuit.
Then resistance $2 Ω$ and $4 Ω$ are in series connection.
$∴ R_{eq} = 2 + 4 = 6 Ω$
Now, current in $4 Ω$ resistance is
$I = \frac{V}{R_{e q}}$
$I = \frac{12}{6} = 2 A$

JCECE-2017]![original image](https://images.teachhire.in/YCT%20PHYSICS/14.%20SEMICONDUCTOR%20ELECTRONICS-%20MATERIALS-%20DEVICES%20AND%20SIMPLE%20CIRCUITS/CLFBqnRIYnI4klcFUltGCy-E1jk4q0hhda23jWA6G7Y.original.fullsize.png)

Semiconductor Electronics Material Devices and Simple Circuits

150752 A diode is connected to $220 V$ (rms) $AC$ in series with a capacitor as shown in figure. The voltage across the capacitor is
original image

1

220 V

2

110 V

3

311.1 V

4

\frac{220}{\sqrt{2}} V

Explanation:

D In this circuit diode D conduct in positive half cycle and in negative half cycle this is open circuit so it behaves as half wave rectifier.
The voltage across the capacitor,
$V_{rms} = \frac{V_{0}}{2} = \frac{220 \sqrt{2}}{2} \Rightarrow \frac{220}{\sqrt{2}}$

JCECE-2012

Semiconductor Electronics Material Devices and Simple Circuits

150761 In a transistor connected in a common emitter $mode, R_{C} = 4 k Ω, R_{1} = 1 k Ω, I_{C} = 1 mA$ and $I_{b} =$ $20 μ A$ . The voltage gain is

1 100

2 200

3 300

4 400

Explanation:

B Given that,
$R_{c} = 4 \times 10^{3} Ω R_{1} = 1 \times 10^{3} Ω$
$I_{C} = 1 \times 10^{- 3} I_{B} = 20 \times 10^{- 6} A$
We know that, voltage gain,
$A_{V} = \frac{β R_{C}}{R_{1}} = \frac{I_{C}}{I_{B}} \frac{R_{c}}{R_{1}} (∵ β = \frac{I_{C}}{I_{B}})$
$A_{v} = \frac{1 \times 10^{- 3} \times 4 \times 10^{3}}{20 \times 10^{- 6} \times 1 \times 10^{3}}$
$A_{V} = 200$

COMEDK 2015

Semiconductor Electronics Material Devices and Simple Circuits

150749 For the given circuit shown below, to act as full wave rectifier, the $AC$ input should be connected across and and the DC. output would appear across and
original image

1 B and D and A and C

2 B and A and C and D

3 C and A and B and D

4 C and D and B and A

5 none of the above

Explanation:

B Let naming the diodes $D_{1}, D_{2}, D_{3}$ and $D_{4}$
original image
According to the question circuit is act as full wave rectifier.
In above figure, when we apply $AC$ input across $BD$ . For positive half cycle diode $D_{1} & D_{3}$ act as forward bias and $D_{2} & D_{4}$ act as reverse bias.
For negative half cycle $D_{1} & D_{3}$ act as a reverse bias and $D_{2} & D_{4}$ act as forward bias.
Hence, AC input should be connected across B and D and the DC. Output would appear across A and C.

Kerala CEE 2004

Semiconductor Electronics Material Devices and Simple Circuits

150750 In the circuit of figure, treat diode as ideal, current in the $4 Ω$ resistor is

1

2 A

2

3 A

3

\frac{12}{7} A

4

\frac{30}{13} A

Explanation:

B original image
From above figure, we can see diode $D_{1}$ is forward bias so, current flow through it and $D_{2}$ is reverse bias and it acts like a open circuit.
Then resistance $2 Ω$ and $4 Ω$ are in series connection.
$∴ R_{eq} = 2 + 4 = 6 Ω$
Now, current in $4 Ω$ resistance is
$I = \frac{V}{R_{e q}}$
$I = \frac{12}{6} = 2 A$

JCECE-2017]![original image](https://images.teachhire.in/YCT%20PHYSICS/14.%20SEMICONDUCTOR%20ELECTRONICS-%20MATERIALS-%20DEVICES%20AND%20SIMPLE%20CIRCUITS/CLFBqnRIYnI4klcFUltGCy-E1jk4q0hhda23jWA6G7Y.original.fullsize.png)

Semiconductor Electronics Material Devices and Simple Circuits

150752 A diode is connected to $220 V$ (rms) $AC$ in series with a capacitor as shown in figure. The voltage across the capacitor is
original image

1

220 V

2

110 V

3

311.1 V

4

\frac{220}{\sqrt{2}} V

Explanation:

D In this circuit diode D conduct in positive half cycle and in negative half cycle this is open circuit so it behaves as half wave rectifier.
The voltage across the capacitor,
$V_{rms} = \frac{V_{0}}{2} = \frac{220 \sqrt{2}}{2} \Rightarrow \frac{220}{\sqrt{2}}$

JCECE-2012

Semiconductor Electronics Material Devices and Simple Circuits

150761 In a transistor connected in a common emitter $mode, R_{C} = 4 k Ω, R_{1} = 1 k Ω, I_{C} = 1 mA$ and $I_{b} =$ $20 μ A$ . The voltage gain is

1 100

2 200

3 300

4 400

Explanation:

B Given that,
$R_{c} = 4 \times 10^{3} Ω R_{1} = 1 \times 10^{3} Ω$
$I_{C} = 1 \times 10^{- 3} I_{B} = 20 \times 10^{- 6} A$
We know that, voltage gain,
$A_{V} = \frac{β R_{C}}{R_{1}} = \frac{I_{C}}{I_{B}} \frac{R_{c}}{R_{1}} (∵ β = \frac{I_{C}}{I_{B}})$
$A_{v} = \frac{1 \times 10^{- 3} \times 4 \times 10^{3}}{20 \times 10^{- 6} \times 1 \times 10^{3}}$
$A_{V} = 200$

COMEDK 2015

Semiconductor Electronics Material Devices and Simple Circuits

150749 For the given circuit shown below, to act as full wave rectifier, the $AC$ input should be connected across and and the DC. output would appear across and
original image

1 B and D and A and C

2 B and A and C and D

3 C and A and B and D

4 C and D and B and A

5 none of the above

Explanation:

B Let naming the diodes $D_{1}, D_{2}, D_{3}$ and $D_{4}$
original image
According to the question circuit is act as full wave rectifier.
In above figure, when we apply $AC$ input across $BD$ . For positive half cycle diode $D_{1} & D_{3}$ act as forward bias and $D_{2} & D_{4}$ act as reverse bias.
For negative half cycle $D_{1} & D_{3}$ act as a reverse bias and $D_{2} & D_{4}$ act as forward bias.
Hence, AC input should be connected across B and D and the DC. Output would appear across A and C.

Kerala CEE 2004

Semiconductor Electronics Material Devices and Simple Circuits

150750 In the circuit of figure, treat diode as ideal, current in the $4 Ω$ resistor is

1

2 A

2

3 A

3

\frac{12}{7} A

4

\frac{30}{13} A

Explanation:

B original image
From above figure, we can see diode $D_{1}$ is forward bias so, current flow through it and $D_{2}$ is reverse bias and it acts like a open circuit.
Then resistance $2 Ω$ and $4 Ω$ are in series connection.
$∴ R_{eq} = 2 + 4 = 6 Ω$
Now, current in $4 Ω$ resistance is
$I = \frac{V}{R_{e q}}$
$I = \frac{12}{6} = 2 A$

JCECE-2017]![original image](https://images.teachhire.in/YCT%20PHYSICS/14.%20SEMICONDUCTOR%20ELECTRONICS-%20MATERIALS-%20DEVICES%20AND%20SIMPLE%20CIRCUITS/CLFBqnRIYnI4klcFUltGCy-E1jk4q0hhda23jWA6G7Y.original.fullsize.png)

Semiconductor Electronics Material Devices and Simple Circuits

150752 A diode is connected to $220 V$ (rms) $AC$ in series with a capacitor as shown in figure. The voltage across the capacitor is
original image

1

220 V

2

110 V

3

311.1 V

4

\frac{220}{\sqrt{2}} V

Explanation:

D In this circuit diode D conduct in positive half cycle and in negative half cycle this is open circuit so it behaves as half wave rectifier.
The voltage across the capacitor,
$V_{rms} = \frac{V_{0}}{2} = \frac{220 \sqrt{2}}{2} \Rightarrow \frac{220}{\sqrt{2}}$

JCECE-2012

Semiconductor Electronics Material Devices and Simple Circuits

150761 In a transistor connected in a common emitter $mode, R_{C} = 4 k Ω, R_{1} = 1 k Ω, I_{C} = 1 mA$ and $I_{b} =$ $20 μ A$ . The voltage gain is

1 100

2 200

3 300

4 400

Explanation:

B Given that,
$R_{c} = 4 \times 10^{3} Ω R_{1} = 1 \times 10^{3} Ω$
$I_{C} = 1 \times 10^{- 3} I_{B} = 20 \times 10^{- 6} A$
We know that, voltage gain,
$A_{V} = \frac{β R_{C}}{R_{1}} = \frac{I_{C}}{I_{B}} \frac{R_{c}}{R_{1}} (∵ β = \frac{I_{C}}{I_{B}})$
$A_{v} = \frac{1 \times 10^{- 3} \times 4 \times 10^{3}}{20 \times 10^{- 6} \times 1 \times 10^{3}}$
$A_{V} = 200$

COMEDK 2015