QBManager

Semiconductor Electronics Material Devices and Simple Circuits

150906 Consider the junction diode as ideal. The value of current flowing through $A B$ is
original image

1

10^{- 2} A

2

10^{- 1} A

3

10^{- 3} A

4

0 A

Explanation:

B original image
We know that,
$V = R I$
$V_{A} - V_{B} = I R$
$4 - (- 6) = 1000 \times I$
$\frac{10}{1000} = I$
$I = 10^{- 2} A$

NEET- 2016

Semiconductor Electronics Material Devices and Simple Circuits

150907 The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R_{1}$ will be
original image

1

2.5 A

2

10.0 A

3

1.43 A

4

3.13 A

Explanation:

B Given,
Diode $D_{1}$ is reverse bias so current will not flow.
Diode $D_{2}$ is forward bias so it will flow the current.
So, $I = \frac{10}{4} = 2.5 A$

NEET- 2016

Semiconductor Electronics Material Devices and Simple Circuits

150909 In the given figure, a diode $D$ is connected to an external resistance $R = 100 Ω$ and an e.m.f of 3.5 $V$ . If the barrier potential developed across the diode is $0.5 V$ , the current in the circuit will be
original image

1

30 mA

2

40 mA

3

20 mA

4

35 mA

Explanation:

B Given,
External resistance $(R) = 100 R$
$emf = 3.5 V$
Barrier Potential $(V_{barrier}) = 0.5 V$
Current (I) =?
Net potential drop $(V_{net}) = emf - V_{barrier}$
$= 3.5 - 0.5$
$V_{net} = 3 V$
So,
$I = \frac{3}{100} = 30 \times 10^{- 3} A$
$I = 30 mA$

AIPMT- 2015

Semiconductor Electronics Material Devices and Simple Circuits

150910 The given graph represents $V - I$ characteristic for a semiconductor device. Which of the following statement is correct?
original image

1 It is V-I characteristic for solar cell where point

A

represents open circuit voltage and point B short circuit current

2 It is for a solar cell and points A and B represent open circuit voltage and current, respectively

3 It is for a photodiode and points A and B represent open circuit voltage and current, respectively

4 It is for a LED and points A and B represent open circuit voltage and short circuit current respectively

Explanation:

B In V-I characteristic for solar cell where point A represents open circuit voltage and point B short circuit current.
original image
At point B
open circuit voltage $(V_{oc}) = 0$
at point $A$
short circuit current $(I_{sc}) = I$
Short circuit current $I = I$
potential $V = 0$

AIPMT- 2014

Semiconductor Electronics Material Devices and Simple Circuits

150853 The circuit shown in the figure contains two diodes each with a forward resistance of $50 Ω$ and with infinite backward resistance if the battery is $6 V$ , the current through the $100 Ω$ resistance (in) ampere-
original image

1 0

2

0.02 A

3

0.03 A

4

0.036 A

Explanation:

B Given That,
$6 V$
original image
Diode $D_{2}$ is in reverse bias so it will be open circuit. So, new circuit.

$∴ diode has a resistance of r_{d} = 50 Ω$
$∵ I = \frac{6 V}{150 + 100 + 50} = \frac{6}{300} = 2 \times 10^{- 2} = 0.02 A$
$I = 0.02 A$

SRMJEEE - 2008

Semiconductor Electronics Material Devices and Simple Circuits

150906 Consider the junction diode as ideal. The value of current flowing through $A B$ is
original image

1

10^{- 2} A

2

10^{- 1} A

3

10^{- 3} A

4

0 A

Explanation:

B original image
We know that,
$V = R I$
$V_{A} - V_{B} = I R$
$4 - (- 6) = 1000 \times I$
$\frac{10}{1000} = I$
$I = 10^{- 2} A$

NEET- 2016

Semiconductor Electronics Material Devices and Simple Circuits

150907 The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R_{1}$ will be
original image

1

2.5 A

2

10.0 A

3

1.43 A

4

3.13 A

Explanation:

B Given,
Diode $D_{1}$ is reverse bias so current will not flow.
Diode $D_{2}$ is forward bias so it will flow the current.
So, $I = \frac{10}{4} = 2.5 A$

NEET- 2016

Semiconductor Electronics Material Devices and Simple Circuits

150909 In the given figure, a diode $D$ is connected to an external resistance $R = 100 Ω$ and an e.m.f of 3.5 $V$ . If the barrier potential developed across the diode is $0.5 V$ , the current in the circuit will be
original image

1

30 mA

2

40 mA

3

20 mA

4

35 mA

Explanation:

B Given,
External resistance $(R) = 100 R$
$emf = 3.5 V$
Barrier Potential $(V_{barrier}) = 0.5 V$
Current (I) =?
Net potential drop $(V_{net}) = emf - V_{barrier}$
$= 3.5 - 0.5$
$V_{net} = 3 V$
So,
$I = \frac{3}{100} = 30 \times 10^{- 3} A$
$I = 30 mA$

AIPMT- 2015

Semiconductor Electronics Material Devices and Simple Circuits

150910 The given graph represents $V - I$ characteristic for a semiconductor device. Which of the following statement is correct?
original image

1 It is V-I characteristic for solar cell where point

A

represents open circuit voltage and point B short circuit current

2 It is for a solar cell and points A and B represent open circuit voltage and current, respectively

3 It is for a photodiode and points A and B represent open circuit voltage and current, respectively

4 It is for a LED and points A and B represent open circuit voltage and short circuit current respectively

Explanation:

B In V-I characteristic for solar cell where point A represents open circuit voltage and point B short circuit current.
original image
At point B
open circuit voltage $(V_{oc}) = 0$
at point $A$
short circuit current $(I_{sc}) = I$
Short circuit current $I = I$
potential $V = 0$

AIPMT- 2014

Semiconductor Electronics Material Devices and Simple Circuits

150853 The circuit shown in the figure contains two diodes each with a forward resistance of $50 Ω$ and with infinite backward resistance if the battery is $6 V$ , the current through the $100 Ω$ resistance (in) ampere-
original image

1 0

2

0.02 A

3

0.03 A

4

0.036 A

Explanation:

B Given That,
$6 V$
original image
Diode $D_{2}$ is in reverse bias so it will be open circuit. So, new circuit.

$∴ diode has a resistance of r_{d} = 50 Ω$
$∵ I = \frac{6 V}{150 + 100 + 50} = \frac{6}{300} = 2 \times 10^{- 2} = 0.02 A$
$I = 0.02 A$

SRMJEEE - 2008

Semiconductor Electronics Material Devices and Simple Circuits

150906 Consider the junction diode as ideal. The value of current flowing through $A B$ is
original image

1

10^{- 2} A

2

10^{- 1} A

3

10^{- 3} A

4

0 A

Explanation:

B original image
We know that,
$V = R I$
$V_{A} - V_{B} = I R$
$4 - (- 6) = 1000 \times I$
$\frac{10}{1000} = I$
$I = 10^{- 2} A$

NEET- 2016

Semiconductor Electronics Material Devices and Simple Circuits

150907 The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R_{1}$ will be
original image

1

2.5 A

2

10.0 A

3

1.43 A

4

3.13 A

Explanation:

B Given,
Diode $D_{1}$ is reverse bias so current will not flow.
Diode $D_{2}$ is forward bias so it will flow the current.
So, $I = \frac{10}{4} = 2.5 A$

NEET- 2016

Semiconductor Electronics Material Devices and Simple Circuits

150909 In the given figure, a diode $D$ is connected to an external resistance $R = 100 Ω$ and an e.m.f of 3.5 $V$ . If the barrier potential developed across the diode is $0.5 V$ , the current in the circuit will be
original image

1

30 mA

2

40 mA

3

20 mA

4

35 mA

Explanation:

B Given,
External resistance $(R) = 100 R$
$emf = 3.5 V$
Barrier Potential $(V_{barrier}) = 0.5 V$
Current (I) =?
Net potential drop $(V_{net}) = emf - V_{barrier}$
$= 3.5 - 0.5$
$V_{net} = 3 V$
So,
$I = \frac{3}{100} = 30 \times 10^{- 3} A$
$I = 30 mA$

AIPMT- 2015

Semiconductor Electronics Material Devices and Simple Circuits

150910 The given graph represents $V - I$ characteristic for a semiconductor device. Which of the following statement is correct?
original image

1 It is V-I characteristic for solar cell where point

A

represents open circuit voltage and point B short circuit current

2 It is for a solar cell and points A and B represent open circuit voltage and current, respectively

3 It is for a photodiode and points A and B represent open circuit voltage and current, respectively

4 It is for a LED and points A and B represent open circuit voltage and short circuit current respectively

Explanation:

B In V-I characteristic for solar cell where point A represents open circuit voltage and point B short circuit current.
original image
At point B
open circuit voltage $(V_{oc}) = 0$
at point $A$
short circuit current $(I_{sc}) = I$
Short circuit current $I = I$
potential $V = 0$

AIPMT- 2014

Semiconductor Electronics Material Devices and Simple Circuits

150853 The circuit shown in the figure contains two diodes each with a forward resistance of $50 Ω$ and with infinite backward resistance if the battery is $6 V$ , the current through the $100 Ω$ resistance (in) ampere-
original image

1 0

2

0.02 A

3

0.03 A

4

0.036 A

Explanation:

B Given That,
$6 V$
original image
Diode $D_{2}$ is in reverse bias so it will be open circuit. So, new circuit.

$∴ diode has a resistance of r_{d} = 50 Ω$
$∵ I = \frac{6 V}{150 + 100 + 50} = \frac{6}{300} = 2 \times 10^{- 2} = 0.02 A$
$I = 0.02 A$

SRMJEEE - 2008

Semiconductor Electronics Material Devices and Simple Circuits

150906 Consider the junction diode as ideal. The value of current flowing through $A B$ is
original image

1

10^{- 2} A

2

10^{- 1} A

3

10^{- 3} A

4

0 A

Explanation:

B original image
We know that,
$V = R I$
$V_{A} - V_{B} = I R$
$4 - (- 6) = 1000 \times I$
$\frac{10}{1000} = I$
$I = 10^{- 2} A$

NEET- 2016

Semiconductor Electronics Material Devices and Simple Circuits

150907 The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R_{1}$ will be
original image

1

2.5 A

2

10.0 A

3

1.43 A

4

3.13 A

Explanation:

B Given,
Diode $D_{1}$ is reverse bias so current will not flow.
Diode $D_{2}$ is forward bias so it will flow the current.
So, $I = \frac{10}{4} = 2.5 A$

NEET- 2016

Semiconductor Electronics Material Devices and Simple Circuits

150909 In the given figure, a diode $D$ is connected to an external resistance $R = 100 Ω$ and an e.m.f of 3.5 $V$ . If the barrier potential developed across the diode is $0.5 V$ , the current in the circuit will be
original image

1

30 mA

2

40 mA

3

20 mA

4

35 mA

Explanation:

B Given,
External resistance $(R) = 100 R$
$emf = 3.5 V$
Barrier Potential $(V_{barrier}) = 0.5 V$
Current (I) =?
Net potential drop $(V_{net}) = emf - V_{barrier}$
$= 3.5 - 0.5$
$V_{net} = 3 V$
So,
$I = \frac{3}{100} = 30 \times 10^{- 3} A$
$I = 30 mA$

AIPMT- 2015

Semiconductor Electronics Material Devices and Simple Circuits

150910 The given graph represents $V - I$ characteristic for a semiconductor device. Which of the following statement is correct?
original image

1 It is V-I characteristic for solar cell where point

A

represents open circuit voltage and point B short circuit current

2 It is for a solar cell and points A and B represent open circuit voltage and current, respectively

3 It is for a photodiode and points A and B represent open circuit voltage and current, respectively

4 It is for a LED and points A and B represent open circuit voltage and short circuit current respectively

Explanation:

B In V-I characteristic for solar cell where point A represents open circuit voltage and point B short circuit current.
original image
At point B
open circuit voltage $(V_{oc}) = 0$
at point $A$
short circuit current $(I_{sc}) = I$
Short circuit current $I = I$
potential $V = 0$

AIPMT- 2014

Semiconductor Electronics Material Devices and Simple Circuits

150853 The circuit shown in the figure contains two diodes each with a forward resistance of $50 Ω$ and with infinite backward resistance if the battery is $6 V$ , the current through the $100 Ω$ resistance (in) ampere-
original image

1 0

2

0.02 A

3

0.03 A

4

0.036 A

Explanation:

B Given That,
$6 V$
original image
Diode $D_{2}$ is in reverse bias so it will be open circuit. So, new circuit.

$∴ diode has a resistance of r_{d} = 50 Ω$
$∵ I = \frac{6 V}{150 + 100 + 50} = \frac{6}{300} = 2 \times 10^{- 2} = 0.02 A$
$I = 0.02 A$

SRMJEEE - 2008

Semiconductor Electronics Material Devices and Simple Circuits

150906 Consider the junction diode as ideal. The value of current flowing through $A B$ is
original image

1

10^{- 2} A

2

10^{- 1} A

3

10^{- 3} A

4

0 A

Explanation:

B original image
We know that,
$V = R I$
$V_{A} - V_{B} = I R$
$4 - (- 6) = 1000 \times I$
$\frac{10}{1000} = I$
$I = 10^{- 2} A$

NEET- 2016

Semiconductor Electronics Material Devices and Simple Circuits

150907 The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R_{1}$ will be
original image

1

2.5 A

2

10.0 A

3

1.43 A

4

3.13 A

Explanation:

B Given,
Diode $D_{1}$ is reverse bias so current will not flow.
Diode $D_{2}$ is forward bias so it will flow the current.
So, $I = \frac{10}{4} = 2.5 A$

NEET- 2016

Semiconductor Electronics Material Devices and Simple Circuits

150909 In the given figure, a diode $D$ is connected to an external resistance $R = 100 Ω$ and an e.m.f of 3.5 $V$ . If the barrier potential developed across the diode is $0.5 V$ , the current in the circuit will be
original image

1

30 mA

2

40 mA

3

20 mA

4

35 mA

Explanation:

B Given,
External resistance $(R) = 100 R$
$emf = 3.5 V$
Barrier Potential $(V_{barrier}) = 0.5 V$
Current (I) =?
Net potential drop $(V_{net}) = emf - V_{barrier}$
$= 3.5 - 0.5$
$V_{net} = 3 V$
So,
$I = \frac{3}{100} = 30 \times 10^{- 3} A$
$I = 30 mA$

AIPMT- 2015

Semiconductor Electronics Material Devices and Simple Circuits

150910 The given graph represents $V - I$ characteristic for a semiconductor device. Which of the following statement is correct?
original image

1 It is V-I characteristic for solar cell where point

A

represents open circuit voltage and point B short circuit current

2 It is for a solar cell and points A and B represent open circuit voltage and current, respectively

3 It is for a photodiode and points A and B represent open circuit voltage and current, respectively

4 It is for a LED and points A and B represent open circuit voltage and short circuit current respectively

Explanation:

B In V-I characteristic for solar cell where point A represents open circuit voltage and point B short circuit current.
original image
At point B
open circuit voltage $(V_{oc}) = 0$
at point $A$
short circuit current $(I_{sc}) = I$
Short circuit current $I = I$
potential $V = 0$

AIPMT- 2014

Semiconductor Electronics Material Devices and Simple Circuits

150853 The circuit shown in the figure contains two diodes each with a forward resistance of $50 Ω$ and with infinite backward resistance if the battery is $6 V$ , the current through the $100 Ω$ resistance (in) ampere-
original image

1 0

2

0.02 A

3

0.03 A

4

0.036 A

Explanation:

B Given That,
$6 V$
original image
Diode $D_{2}$ is in reverse bias so it will be open circuit. So, new circuit.

$∴ diode has a resistance of r_{d} = 50 Ω$
$∵ I = \frac{6 V}{150 + 100 + 50} = \frac{6}{300} = 2 \times 10^{- 2} = 0.02 A$
$I = 0.02 A$

SRMJEEE - 2008

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