Semiconductor Electronics Material Devices and Simple Circuits
150733
If the diodes are ideal in the circuit given below, then the current through the cell is
1 \(4 \mathrm{~A}\)
2 \(1.5 \mathrm{~A}\)
3 \(2 \mathrm{~A}\)
4 \(3 \mathrm{~A}\)
Explanation:
C Here Diode \(\mathrm{D}_1\) work as a reverse biased and Diode \(\mathrm{D}_2\) work as a forward biased. So, the circuit simplified as the series connection of two, set of the \(3 \Omega\) and \(2 \Omega\) resistance so the equivalent resistance is, \(\mathrm{R}_{\mathrm{eq}}=2+3+3+2=10 \Omega\) Then, current flowing in circuit, \(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{20}{10}=2 \mathrm{~A}\)
AP EAMCET (22.04.2018) Shift-1]
Semiconductor Electronics Material Devices and Simple Circuits
150734
What is the voltage across an ideal \(p-n\) junction diode for shown circuit?
1 \(0.7 \mathrm{~V}\)
2 \(1 \mathrm{~V}\)
3 \(2 \mathrm{~V}\)
4 \(0 \mathrm{~V}\)
Explanation:
C If the diode is ideal, Then diode act as a reverse biased because the positive terminal of diode is connected to negative terminal of battery so diode act as open circuit. Therefore total voltage drop in diode. Total voltage across the diode will be \(2 \mathrm{~V}\).
UPSEE - 2017
Semiconductor Electronics Material Devices and Simple Circuits
150735
Shown below is a graph of current verses applied voltage for a diode. Approximately what is the resistance of the diode for an applied voltage of \(-1.5 \mathrm{~V}\) ?
1 \(2 \Omega\)
2 \(\infty\)
3 Zero
4 \(1 \Omega\)
Explanation:
B Given, \(\mathrm{V}=-1.5\) volt Then, \(\mathrm{I}=0\) Therefore, Resistance \(=\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{1.5}{0}\) \(\mathrm{R}=\infty\)
UPSEE - 2016
Semiconductor Electronics Material Devices and Simple Circuits
150741
If an ideal junction diode is connected as shown, then the value of the current \(i\) is
1 \(0.013 \mathrm{~A}\)
2 \(0.02 \mathrm{~A}\)
3 \(0.01 \mathrm{~A}\)
4 \(0.1 \mathrm{~A}\)
5 \(0.2 \mathrm{~A}\)
Explanation:
C Given here, Now, potential difference \(\mathrm{V}=3-1=2 \mathrm{~V}\) As we know, \(V=I R\) \(2=I \times 200\) \(I=\frac{2}{200}=0.01 \mathrm{~A}\)
Kerala CEE- 2013]
Semiconductor Electronics Material Devices and Simple Circuits
150743
In the given circuit the current through the battery is
1 \(0.5 \mathrm{~A}\)
2 \(1 \mathrm{~A}\)
3 \(1.5 \mathrm{~A}\)
4 \(2 \mathrm{~A}\)
5 \(2.5 \mathrm{~A}\)
Explanation:
C In the given circuit diode \(\mathrm{D}_1\) is reverse biased so its become open circuit and diode \(\mathrm{D}_2\) and \(\mathrm{D}_3\) are forward biased. So, they will conduct. Thus the circuit becomes - \(5 \Omega\) and \(5 \Omega\) resistance connected in series then equivalent \(\mathrm{R}_{\text {series }}=5 \Omega+5 \Omega=10 \Omega\) Now, \(\quad 10 \Omega\) and \(20 \Omega\) resistance connected in parallel, Therefore, \(\mathrm{R}_{\mathrm{eq}}=\frac{10 \times 20}{10+20}=\frac{10 \times 20}{30}=\frac{20}{3} \Omega\)The current, \(I=\frac{V}{R_{\text {eq }}}=\frac{10}{\frac{20}{3}}=\frac{30}{20}=1.5 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150733
If the diodes are ideal in the circuit given below, then the current through the cell is
1 \(4 \mathrm{~A}\)
2 \(1.5 \mathrm{~A}\)
3 \(2 \mathrm{~A}\)
4 \(3 \mathrm{~A}\)
Explanation:
C Here Diode \(\mathrm{D}_1\) work as a reverse biased and Diode \(\mathrm{D}_2\) work as a forward biased. So, the circuit simplified as the series connection of two, set of the \(3 \Omega\) and \(2 \Omega\) resistance so the equivalent resistance is, \(\mathrm{R}_{\mathrm{eq}}=2+3+3+2=10 \Omega\) Then, current flowing in circuit, \(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{20}{10}=2 \mathrm{~A}\)
AP EAMCET (22.04.2018) Shift-1]
Semiconductor Electronics Material Devices and Simple Circuits
150734
What is the voltage across an ideal \(p-n\) junction diode for shown circuit?
1 \(0.7 \mathrm{~V}\)
2 \(1 \mathrm{~V}\)
3 \(2 \mathrm{~V}\)
4 \(0 \mathrm{~V}\)
Explanation:
C If the diode is ideal, Then diode act as a reverse biased because the positive terminal of diode is connected to negative terminal of battery so diode act as open circuit. Therefore total voltage drop in diode. Total voltage across the diode will be \(2 \mathrm{~V}\).
UPSEE - 2017
Semiconductor Electronics Material Devices and Simple Circuits
150735
Shown below is a graph of current verses applied voltage for a diode. Approximately what is the resistance of the diode for an applied voltage of \(-1.5 \mathrm{~V}\) ?
1 \(2 \Omega\)
2 \(\infty\)
3 Zero
4 \(1 \Omega\)
Explanation:
B Given, \(\mathrm{V}=-1.5\) volt Then, \(\mathrm{I}=0\) Therefore, Resistance \(=\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{1.5}{0}\) \(\mathrm{R}=\infty\)
UPSEE - 2016
Semiconductor Electronics Material Devices and Simple Circuits
150741
If an ideal junction diode is connected as shown, then the value of the current \(i\) is
1 \(0.013 \mathrm{~A}\)
2 \(0.02 \mathrm{~A}\)
3 \(0.01 \mathrm{~A}\)
4 \(0.1 \mathrm{~A}\)
5 \(0.2 \mathrm{~A}\)
Explanation:
C Given here, Now, potential difference \(\mathrm{V}=3-1=2 \mathrm{~V}\) As we know, \(V=I R\) \(2=I \times 200\) \(I=\frac{2}{200}=0.01 \mathrm{~A}\)
Kerala CEE- 2013]
Semiconductor Electronics Material Devices and Simple Circuits
150743
In the given circuit the current through the battery is
1 \(0.5 \mathrm{~A}\)
2 \(1 \mathrm{~A}\)
3 \(1.5 \mathrm{~A}\)
4 \(2 \mathrm{~A}\)
5 \(2.5 \mathrm{~A}\)
Explanation:
C In the given circuit diode \(\mathrm{D}_1\) is reverse biased so its become open circuit and diode \(\mathrm{D}_2\) and \(\mathrm{D}_3\) are forward biased. So, they will conduct. Thus the circuit becomes - \(5 \Omega\) and \(5 \Omega\) resistance connected in series then equivalent \(\mathrm{R}_{\text {series }}=5 \Omega+5 \Omega=10 \Omega\) Now, \(\quad 10 \Omega\) and \(20 \Omega\) resistance connected in parallel, Therefore, \(\mathrm{R}_{\mathrm{eq}}=\frac{10 \times 20}{10+20}=\frac{10 \times 20}{30}=\frac{20}{3} \Omega\)The current, \(I=\frac{V}{R_{\text {eq }}}=\frac{10}{\frac{20}{3}}=\frac{30}{20}=1.5 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150733
If the diodes are ideal in the circuit given below, then the current through the cell is
1 \(4 \mathrm{~A}\)
2 \(1.5 \mathrm{~A}\)
3 \(2 \mathrm{~A}\)
4 \(3 \mathrm{~A}\)
Explanation:
C Here Diode \(\mathrm{D}_1\) work as a reverse biased and Diode \(\mathrm{D}_2\) work as a forward biased. So, the circuit simplified as the series connection of two, set of the \(3 \Omega\) and \(2 \Omega\) resistance so the equivalent resistance is, \(\mathrm{R}_{\mathrm{eq}}=2+3+3+2=10 \Omega\) Then, current flowing in circuit, \(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{20}{10}=2 \mathrm{~A}\)
AP EAMCET (22.04.2018) Shift-1]
Semiconductor Electronics Material Devices and Simple Circuits
150734
What is the voltage across an ideal \(p-n\) junction diode for shown circuit?
1 \(0.7 \mathrm{~V}\)
2 \(1 \mathrm{~V}\)
3 \(2 \mathrm{~V}\)
4 \(0 \mathrm{~V}\)
Explanation:
C If the diode is ideal, Then diode act as a reverse biased because the positive terminal of diode is connected to negative terminal of battery so diode act as open circuit. Therefore total voltage drop in diode. Total voltage across the diode will be \(2 \mathrm{~V}\).
UPSEE - 2017
Semiconductor Electronics Material Devices and Simple Circuits
150735
Shown below is a graph of current verses applied voltage for a diode. Approximately what is the resistance of the diode for an applied voltage of \(-1.5 \mathrm{~V}\) ?
1 \(2 \Omega\)
2 \(\infty\)
3 Zero
4 \(1 \Omega\)
Explanation:
B Given, \(\mathrm{V}=-1.5\) volt Then, \(\mathrm{I}=0\) Therefore, Resistance \(=\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{1.5}{0}\) \(\mathrm{R}=\infty\)
UPSEE - 2016
Semiconductor Electronics Material Devices and Simple Circuits
150741
If an ideal junction diode is connected as shown, then the value of the current \(i\) is
1 \(0.013 \mathrm{~A}\)
2 \(0.02 \mathrm{~A}\)
3 \(0.01 \mathrm{~A}\)
4 \(0.1 \mathrm{~A}\)
5 \(0.2 \mathrm{~A}\)
Explanation:
C Given here, Now, potential difference \(\mathrm{V}=3-1=2 \mathrm{~V}\) As we know, \(V=I R\) \(2=I \times 200\) \(I=\frac{2}{200}=0.01 \mathrm{~A}\)
Kerala CEE- 2013]
Semiconductor Electronics Material Devices and Simple Circuits
150743
In the given circuit the current through the battery is
1 \(0.5 \mathrm{~A}\)
2 \(1 \mathrm{~A}\)
3 \(1.5 \mathrm{~A}\)
4 \(2 \mathrm{~A}\)
5 \(2.5 \mathrm{~A}\)
Explanation:
C In the given circuit diode \(\mathrm{D}_1\) is reverse biased so its become open circuit and diode \(\mathrm{D}_2\) and \(\mathrm{D}_3\) are forward biased. So, they will conduct. Thus the circuit becomes - \(5 \Omega\) and \(5 \Omega\) resistance connected in series then equivalent \(\mathrm{R}_{\text {series }}=5 \Omega+5 \Omega=10 \Omega\) Now, \(\quad 10 \Omega\) and \(20 \Omega\) resistance connected in parallel, Therefore, \(\mathrm{R}_{\mathrm{eq}}=\frac{10 \times 20}{10+20}=\frac{10 \times 20}{30}=\frac{20}{3} \Omega\)The current, \(I=\frac{V}{R_{\text {eq }}}=\frac{10}{\frac{20}{3}}=\frac{30}{20}=1.5 \mathrm{~A}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Semiconductor Electronics Material Devices and Simple Circuits
150733
If the diodes are ideal in the circuit given below, then the current through the cell is
1 \(4 \mathrm{~A}\)
2 \(1.5 \mathrm{~A}\)
3 \(2 \mathrm{~A}\)
4 \(3 \mathrm{~A}\)
Explanation:
C Here Diode \(\mathrm{D}_1\) work as a reverse biased and Diode \(\mathrm{D}_2\) work as a forward biased. So, the circuit simplified as the series connection of two, set of the \(3 \Omega\) and \(2 \Omega\) resistance so the equivalent resistance is, \(\mathrm{R}_{\mathrm{eq}}=2+3+3+2=10 \Omega\) Then, current flowing in circuit, \(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{20}{10}=2 \mathrm{~A}\)
AP EAMCET (22.04.2018) Shift-1]
Semiconductor Electronics Material Devices and Simple Circuits
150734
What is the voltage across an ideal \(p-n\) junction diode for shown circuit?
1 \(0.7 \mathrm{~V}\)
2 \(1 \mathrm{~V}\)
3 \(2 \mathrm{~V}\)
4 \(0 \mathrm{~V}\)
Explanation:
C If the diode is ideal, Then diode act as a reverse biased because the positive terminal of diode is connected to negative terminal of battery so diode act as open circuit. Therefore total voltage drop in diode. Total voltage across the diode will be \(2 \mathrm{~V}\).
UPSEE - 2017
Semiconductor Electronics Material Devices and Simple Circuits
150735
Shown below is a graph of current verses applied voltage for a diode. Approximately what is the resistance of the diode for an applied voltage of \(-1.5 \mathrm{~V}\) ?
1 \(2 \Omega\)
2 \(\infty\)
3 Zero
4 \(1 \Omega\)
Explanation:
B Given, \(\mathrm{V}=-1.5\) volt Then, \(\mathrm{I}=0\) Therefore, Resistance \(=\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{1.5}{0}\) \(\mathrm{R}=\infty\)
UPSEE - 2016
Semiconductor Electronics Material Devices and Simple Circuits
150741
If an ideal junction diode is connected as shown, then the value of the current \(i\) is
1 \(0.013 \mathrm{~A}\)
2 \(0.02 \mathrm{~A}\)
3 \(0.01 \mathrm{~A}\)
4 \(0.1 \mathrm{~A}\)
5 \(0.2 \mathrm{~A}\)
Explanation:
C Given here, Now, potential difference \(\mathrm{V}=3-1=2 \mathrm{~V}\) As we know, \(V=I R\) \(2=I \times 200\) \(I=\frac{2}{200}=0.01 \mathrm{~A}\)
Kerala CEE- 2013]
Semiconductor Electronics Material Devices and Simple Circuits
150743
In the given circuit the current through the battery is
1 \(0.5 \mathrm{~A}\)
2 \(1 \mathrm{~A}\)
3 \(1.5 \mathrm{~A}\)
4 \(2 \mathrm{~A}\)
5 \(2.5 \mathrm{~A}\)
Explanation:
C In the given circuit diode \(\mathrm{D}_1\) is reverse biased so its become open circuit and diode \(\mathrm{D}_2\) and \(\mathrm{D}_3\) are forward biased. So, they will conduct. Thus the circuit becomes - \(5 \Omega\) and \(5 \Omega\) resistance connected in series then equivalent \(\mathrm{R}_{\text {series }}=5 \Omega+5 \Omega=10 \Omega\) Now, \(\quad 10 \Omega\) and \(20 \Omega\) resistance connected in parallel, Therefore, \(\mathrm{R}_{\mathrm{eq}}=\frac{10 \times 20}{10+20}=\frac{10 \times 20}{30}=\frac{20}{3} \Omega\)The current, \(I=\frac{V}{R_{\text {eq }}}=\frac{10}{\frac{20}{3}}=\frac{30}{20}=1.5 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150733
If the diodes are ideal in the circuit given below, then the current through the cell is
1 \(4 \mathrm{~A}\)
2 \(1.5 \mathrm{~A}\)
3 \(2 \mathrm{~A}\)
4 \(3 \mathrm{~A}\)
Explanation:
C Here Diode \(\mathrm{D}_1\) work as a reverse biased and Diode \(\mathrm{D}_2\) work as a forward biased. So, the circuit simplified as the series connection of two, set of the \(3 \Omega\) and \(2 \Omega\) resistance so the equivalent resistance is, \(\mathrm{R}_{\mathrm{eq}}=2+3+3+2=10 \Omega\) Then, current flowing in circuit, \(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{20}{10}=2 \mathrm{~A}\)
AP EAMCET (22.04.2018) Shift-1]
Semiconductor Electronics Material Devices and Simple Circuits
150734
What is the voltage across an ideal \(p-n\) junction diode for shown circuit?
1 \(0.7 \mathrm{~V}\)
2 \(1 \mathrm{~V}\)
3 \(2 \mathrm{~V}\)
4 \(0 \mathrm{~V}\)
Explanation:
C If the diode is ideal, Then diode act as a reverse biased because the positive terminal of diode is connected to negative terminal of battery so diode act as open circuit. Therefore total voltage drop in diode. Total voltage across the diode will be \(2 \mathrm{~V}\).
UPSEE - 2017
Semiconductor Electronics Material Devices and Simple Circuits
150735
Shown below is a graph of current verses applied voltage for a diode. Approximately what is the resistance of the diode for an applied voltage of \(-1.5 \mathrm{~V}\) ?
1 \(2 \Omega\)
2 \(\infty\)
3 Zero
4 \(1 \Omega\)
Explanation:
B Given, \(\mathrm{V}=-1.5\) volt Then, \(\mathrm{I}=0\) Therefore, Resistance \(=\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{1.5}{0}\) \(\mathrm{R}=\infty\)
UPSEE - 2016
Semiconductor Electronics Material Devices and Simple Circuits
150741
If an ideal junction diode is connected as shown, then the value of the current \(i\) is
1 \(0.013 \mathrm{~A}\)
2 \(0.02 \mathrm{~A}\)
3 \(0.01 \mathrm{~A}\)
4 \(0.1 \mathrm{~A}\)
5 \(0.2 \mathrm{~A}\)
Explanation:
C Given here, Now, potential difference \(\mathrm{V}=3-1=2 \mathrm{~V}\) As we know, \(V=I R\) \(2=I \times 200\) \(I=\frac{2}{200}=0.01 \mathrm{~A}\)
Kerala CEE- 2013]
Semiconductor Electronics Material Devices and Simple Circuits
150743
In the given circuit the current through the battery is
1 \(0.5 \mathrm{~A}\)
2 \(1 \mathrm{~A}\)
3 \(1.5 \mathrm{~A}\)
4 \(2 \mathrm{~A}\)
5 \(2.5 \mathrm{~A}\)
Explanation:
C In the given circuit diode \(\mathrm{D}_1\) is reverse biased so its become open circuit and diode \(\mathrm{D}_2\) and \(\mathrm{D}_3\) are forward biased. So, they will conduct. Thus the circuit becomes - \(5 \Omega\) and \(5 \Omega\) resistance connected in series then equivalent \(\mathrm{R}_{\text {series }}=5 \Omega+5 \Omega=10 \Omega\) Now, \(\quad 10 \Omega\) and \(20 \Omega\) resistance connected in parallel, Therefore, \(\mathrm{R}_{\mathrm{eq}}=\frac{10 \times 20}{10+20}=\frac{10 \times 20}{30}=\frac{20}{3} \Omega\)The current, \(I=\frac{V}{R_{\text {eq }}}=\frac{10}{\frac{20}{3}}=\frac{30}{20}=1.5 \mathrm{~A}\)
