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Semiconductor Electronics Material Devices and Simple Circuits

150744 A full-wave rectifier circuit with an $AC$ input is shown
original image
The output voltage across $R_{L}$ is represented as

1

2

3

4

5

Kerala CEE - 2010

Semiconductor Electronics Material Devices and Simple Circuits

150745 In the diagram, the input $AC$ is across the terminals $A$ and $C$ . The output across B and D is
original image

1 same as the input

2 half wave rectified

3 zero

4 full wave rectified

5 steady DC

Explanation:

D original image
For ( + ve) half cycle $D_{1} & D_{3}$ are in forward bais while $D_{2} & D_{4}$ are in reversed bais
Now (-ve) half cycle $\to D_{1} & D_{3}$ are in reverse bias and $D_{2} & D_{4}$ in forward baised
So, given output B and D is in full wave rectifier.

Kerala CEE - 2009

Semiconductor Electronics Material Devices and Simple Circuits

150746 The current I through $10 Ω$ resistor in the circuit given below is
original image

1

50 mA

2

20 mA

3

40 mA

4

80 mA

5

35 mA

Explanation:

D Here $D_{2}$ is reversed biased while $D_{1}$ is forward biased so current flow only $D_{1}$
Then, $I = \frac{V}{R} = \frac{2}{10 + 15} = \frac{2}{25} = 0.08 A$
$I = 80 mA$

Kerala CEE - 2008

Semiconductor Electronics Material Devices and Simple Circuits

150747 When the forward bias voltage of a diode is changed from $0.6 V$ to $0.7 V$ , the current changes from $5 mA$ to $15 mA$ . Then its forward bias resistance is

1

0.01 Ω

2

0.1 Ω

3

10 Ω

4

100 Ω

5

0.2 Ω

Explanation:

C Given that,
$V_{1} = 0.6 V V_{2} = 0.7 V$
$I_{1} = 5 mA I_{2} = 15 mA$
Then forward biased resistance will be,
$r = \frac{dV}{dI} = \frac{0.7 - 0.6}{15 mA - 5 mA} = \frac{0.1}{10 \times 10^{- 3} A}$
$r = 0.1 \times 10^{2} Ω$
$r = 10 Ω$

Kerala CEE 2007

Semiconductor Electronics Material Devices and Simple Circuits

150744 A full-wave rectifier circuit with an $AC$ input is shown
original image
The output voltage across $R_{L}$ is represented as

1

2

3

4

5

Kerala CEE - 2010

Semiconductor Electronics Material Devices and Simple Circuits

150745 In the diagram, the input $AC$ is across the terminals $A$ and $C$ . The output across B and D is
original image

1 same as the input

2 half wave rectified

3 zero

4 full wave rectified

5 steady DC

Explanation:

D original image
For ( + ve) half cycle $D_{1} & D_{3}$ are in forward bais while $D_{2} & D_{4}$ are in reversed bais
Now (-ve) half cycle $\to D_{1} & D_{3}$ are in reverse bias and $D_{2} & D_{4}$ in forward baised
So, given output B and D is in full wave rectifier.

Kerala CEE - 2009

Semiconductor Electronics Material Devices and Simple Circuits

150746 The current I through $10 Ω$ resistor in the circuit given below is
original image

1

50 mA

2

20 mA

3

40 mA

4

80 mA

5

35 mA

Explanation:

D Here $D_{2}$ is reversed biased while $D_{1}$ is forward biased so current flow only $D_{1}$
Then, $I = \frac{V}{R} = \frac{2}{10 + 15} = \frac{2}{25} = 0.08 A$
$I = 80 mA$

Kerala CEE - 2008

Semiconductor Electronics Material Devices and Simple Circuits

150747 When the forward bias voltage of a diode is changed from $0.6 V$ to $0.7 V$ , the current changes from $5 mA$ to $15 mA$ . Then its forward bias resistance is

1

0.01 Ω

2

0.1 Ω

3

10 Ω

4

100 Ω

5

0.2 Ω

Explanation:

C Given that,
$V_{1} = 0.6 V V_{2} = 0.7 V$
$I_{1} = 5 mA I_{2} = 15 mA$
Then forward biased resistance will be,
$r = \frac{dV}{dI} = \frac{0.7 - 0.6}{15 mA - 5 mA} = \frac{0.1}{10 \times 10^{- 3} A}$
$r = 0.1 \times 10^{2} Ω$
$r = 10 Ω$

Kerala CEE 2007

Semiconductor Electronics Material Devices and Simple Circuits

150744 A full-wave rectifier circuit with an $AC$ input is shown
original image
The output voltage across $R_{L}$ is represented as

1

2

3

4

5

Kerala CEE - 2010

Semiconductor Electronics Material Devices and Simple Circuits

150745 In the diagram, the input $AC$ is across the terminals $A$ and $C$ . The output across B and D is
original image

1 same as the input

2 half wave rectified

3 zero

4 full wave rectified

5 steady DC

Explanation:

D original image
For ( + ve) half cycle $D_{1} & D_{3}$ are in forward bais while $D_{2} & D_{4}$ are in reversed bais
Now (-ve) half cycle $\to D_{1} & D_{3}$ are in reverse bias and $D_{2} & D_{4}$ in forward baised
So, given output B and D is in full wave rectifier.

Kerala CEE - 2009

Semiconductor Electronics Material Devices and Simple Circuits

150746 The current I through $10 Ω$ resistor in the circuit given below is
original image

1

50 mA

2

20 mA

3

40 mA

4

80 mA

5

35 mA

Explanation:

D Here $D_{2}$ is reversed biased while $D_{1}$ is forward biased so current flow only $D_{1}$
Then, $I = \frac{V}{R} = \frac{2}{10 + 15} = \frac{2}{25} = 0.08 A$
$I = 80 mA$

Kerala CEE - 2008

Semiconductor Electronics Material Devices and Simple Circuits

150747 When the forward bias voltage of a diode is changed from $0.6 V$ to $0.7 V$ , the current changes from $5 mA$ to $15 mA$ . Then its forward bias resistance is

1

0.01 Ω

2

0.1 Ω

3

10 Ω

4

100 Ω

5

0.2 Ω

Explanation:

C Given that,
$V_{1} = 0.6 V V_{2} = 0.7 V$
$I_{1} = 5 mA I_{2} = 15 mA$
Then forward biased resistance will be,
$r = \frac{dV}{dI} = \frac{0.7 - 0.6}{15 mA - 5 mA} = \frac{0.1}{10 \times 10^{- 3} A}$
$r = 0.1 \times 10^{2} Ω$
$r = 10 Ω$

Kerala CEE 2007

Semiconductor Electronics Material Devices and Simple Circuits

150744 A full-wave rectifier circuit with an $AC$ input is shown
original image
The output voltage across $R_{L}$ is represented as

1

2

3

4

5

Kerala CEE - 2010

Semiconductor Electronics Material Devices and Simple Circuits

150745 In the diagram, the input $AC$ is across the terminals $A$ and $C$ . The output across B and D is
original image

1 same as the input

2 half wave rectified

3 zero

4 full wave rectified

5 steady DC

Explanation:

D original image
For ( + ve) half cycle $D_{1} & D_{3}$ are in forward bais while $D_{2} & D_{4}$ are in reversed bais
Now (-ve) half cycle $\to D_{1} & D_{3}$ are in reverse bias and $D_{2} & D_{4}$ in forward baised
So, given output B and D is in full wave rectifier.

Kerala CEE - 2009

Semiconductor Electronics Material Devices and Simple Circuits

150746 The current I through $10 Ω$ resistor in the circuit given below is
original image

1

50 mA

2

20 mA

3

40 mA

4

80 mA

5

35 mA

Explanation:

D Here $D_{2}$ is reversed biased while $D_{1}$ is forward biased so current flow only $D_{1}$
Then, $I = \frac{V}{R} = \frac{2}{10 + 15} = \frac{2}{25} = 0.08 A$
$I = 80 mA$

Kerala CEE - 2008

Semiconductor Electronics Material Devices and Simple Circuits

150747 When the forward bias voltage of a diode is changed from $0.6 V$ to $0.7 V$ , the current changes from $5 mA$ to $15 mA$ . Then its forward bias resistance is

1

0.01 Ω

2

0.1 Ω

3

10 Ω

4

100 Ω

5

0.2 Ω

Explanation:

C Given that,
$V_{1} = 0.6 V V_{2} = 0.7 V$
$I_{1} = 5 mA I_{2} = 15 mA$
Then forward biased resistance will be,
$r = \frac{dV}{dI} = \frac{0.7 - 0.6}{15 mA - 5 mA} = \frac{0.1}{10 \times 10^{- 3} A}$
$r = 0.1 \times 10^{2} Ω$
$r = 10 Ω$

Kerala CEE 2007

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