Semiconductor Electronics Material Devices and Simple Circuits
150764
The current I through \(\mathbf{1 0} \mathrm{ohm}\) resistor in the circuit given below is
1 \(50 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(80 \mathrm{~mA}\)
Explanation:
D In this circuit \(\mathrm{D}_2\) is reverse biased so, no current flow across \(\mathrm{D}_2\), now circuit become . Then, \(i=\frac{V}{R_{\text {eq }}}=\frac{2}{10+15}\) \(i=\frac{2}{25}=0.08 \mathrm{~A}\) \(i=80 \mathrm{~mA}\)
COMEDK 2019
Semiconductor Electronics Material Devices and Simple Circuits
150765
Assertion: The value of current through \(p-n\) junction in the given figure will be \(10 \mathrm{~mA}\). Reason: In the above figure, \(p\)-side is at higher potential than n-side.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B The current, \(I=\frac{V}{R}=\frac{5-2}{300}=\frac{3}{300}=10^{-2} \mathrm{~A}\) \(I=10 \times 10^{-3}\) \(I=10 \mathrm{~mA} \text { (correct) }\) Reason-: \(\mathrm{P}\) side Voltage is \(5 \mathrm{~V}\) which is greater than \(\mathrm{N}\) side voltage is \(+2 \mathrm{~V}\) so correct.
AIIMS-2008
Semiconductor Electronics Material Devices and Simple Circuits
150766
A silicon diode has a threshold voltage of \(0.7 \mathrm{~V}\). If an input voltage given by \(2 \sin (\pi t)\) is supplied to a half wave rectifier circuit using this diode, the rectified output has a peak value.
1 \(2 \mathrm{~V}\)
2 \(14 . \mathrm{V}\)
3 \(1.3 \mathrm{~V}\)
4 \(0.7 \mathrm{~V}\)
Explanation:
C Given, Input voltage \(=2 \sin (\pi t)\) So, \(\quad\) Input peak voltage \(=2 \mathrm{~V}\) Barrier voltage \(=0.7 \mathrm{~V}\) \(\because\) The peak value of rectified output voltage \(=\) Peak voltage - threshold voltage \(=2-0.7\) \(=1.3 \mathrm{~V}\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150767
If current in diode is five times that in \(R_1\) and breakdown voltage of diode is 6 volt then find R.
1 \(\frac{1000}{3} \Omega\)
2 \(\frac{2000}{3} \Omega\)
3 \(\frac{4000}{5} \Omega\)
4 \(\frac{5000}{3} \Omega\)
Explanation:
B Current in diode is 5 times that in \(\mathrm{R}_1\) \(I_1=\frac{V}{R_1}=\frac{6}{1 \times 10^3} A\) \(I_1=6 \mathrm{~m} \mathrm{~A}\) Then, current in the diode \(\mathrm{I}_{\mathrm{D}}=6 \times 5=30 \mathrm{~mA}\) So, total current drawn from battery \(=6 \mathrm{~m} \mathrm{~A}+30 \mathrm{~m} \mathrm{~A} \Rightarrow 36 \mathrm{~m} \mathrm{~A}\) Now potential difference across resistance \(\mathrm{R}\) \(\mathrm{V}=24 \text { volt }\) So, \(\mathrm{V}=\mathrm{IR}\) \(24=36 \times 10^{-3} \mathrm{R}\) \(\mathrm{R}=\frac{2000}{3} \Omega\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150769
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is
1 \(0.75 \mathrm{~A}\)
2 \(0.5 \mathrm{~A}\)
3 \(0.25 \mathrm{~A}\)
4 zero
Explanation:
B In diode \(D_1\) the current pass through \(P\) side so current flow in diode \(D_1\) but in case of diode \(D_2\) there is no current flow because it is reverse bias. So \(\quad \mathrm{I}=\mathrm{V} / \mathrm{R}\) \(I=\frac{5}{10}\) \(I=0.5 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150764
The current I through \(\mathbf{1 0} \mathrm{ohm}\) resistor in the circuit given below is
1 \(50 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(80 \mathrm{~mA}\)
Explanation:
D In this circuit \(\mathrm{D}_2\) is reverse biased so, no current flow across \(\mathrm{D}_2\), now circuit become . Then, \(i=\frac{V}{R_{\text {eq }}}=\frac{2}{10+15}\) \(i=\frac{2}{25}=0.08 \mathrm{~A}\) \(i=80 \mathrm{~mA}\)
COMEDK 2019
Semiconductor Electronics Material Devices and Simple Circuits
150765
Assertion: The value of current through \(p-n\) junction in the given figure will be \(10 \mathrm{~mA}\). Reason: In the above figure, \(p\)-side is at higher potential than n-side.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B The current, \(I=\frac{V}{R}=\frac{5-2}{300}=\frac{3}{300}=10^{-2} \mathrm{~A}\) \(I=10 \times 10^{-3}\) \(I=10 \mathrm{~mA} \text { (correct) }\) Reason-: \(\mathrm{P}\) side Voltage is \(5 \mathrm{~V}\) which is greater than \(\mathrm{N}\) side voltage is \(+2 \mathrm{~V}\) so correct.
AIIMS-2008
Semiconductor Electronics Material Devices and Simple Circuits
150766
A silicon diode has a threshold voltage of \(0.7 \mathrm{~V}\). If an input voltage given by \(2 \sin (\pi t)\) is supplied to a half wave rectifier circuit using this diode, the rectified output has a peak value.
1 \(2 \mathrm{~V}\)
2 \(14 . \mathrm{V}\)
3 \(1.3 \mathrm{~V}\)
4 \(0.7 \mathrm{~V}\)
Explanation:
C Given, Input voltage \(=2 \sin (\pi t)\) So, \(\quad\) Input peak voltage \(=2 \mathrm{~V}\) Barrier voltage \(=0.7 \mathrm{~V}\) \(\because\) The peak value of rectified output voltage \(=\) Peak voltage - threshold voltage \(=2-0.7\) \(=1.3 \mathrm{~V}\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150767
If current in diode is five times that in \(R_1\) and breakdown voltage of diode is 6 volt then find R.
1 \(\frac{1000}{3} \Omega\)
2 \(\frac{2000}{3} \Omega\)
3 \(\frac{4000}{5} \Omega\)
4 \(\frac{5000}{3} \Omega\)
Explanation:
B Current in diode is 5 times that in \(\mathrm{R}_1\) \(I_1=\frac{V}{R_1}=\frac{6}{1 \times 10^3} A\) \(I_1=6 \mathrm{~m} \mathrm{~A}\) Then, current in the diode \(\mathrm{I}_{\mathrm{D}}=6 \times 5=30 \mathrm{~mA}\) So, total current drawn from battery \(=6 \mathrm{~m} \mathrm{~A}+30 \mathrm{~m} \mathrm{~A} \Rightarrow 36 \mathrm{~m} \mathrm{~A}\) Now potential difference across resistance \(\mathrm{R}\) \(\mathrm{V}=24 \text { volt }\) So, \(\mathrm{V}=\mathrm{IR}\) \(24=36 \times 10^{-3} \mathrm{R}\) \(\mathrm{R}=\frac{2000}{3} \Omega\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150769
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is
1 \(0.75 \mathrm{~A}\)
2 \(0.5 \mathrm{~A}\)
3 \(0.25 \mathrm{~A}\)
4 zero
Explanation:
B In diode \(D_1\) the current pass through \(P\) side so current flow in diode \(D_1\) but in case of diode \(D_2\) there is no current flow because it is reverse bias. So \(\quad \mathrm{I}=\mathrm{V} / \mathrm{R}\) \(I=\frac{5}{10}\) \(I=0.5 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150764
The current I through \(\mathbf{1 0} \mathrm{ohm}\) resistor in the circuit given below is
1 \(50 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(80 \mathrm{~mA}\)
Explanation:
D In this circuit \(\mathrm{D}_2\) is reverse biased so, no current flow across \(\mathrm{D}_2\), now circuit become . Then, \(i=\frac{V}{R_{\text {eq }}}=\frac{2}{10+15}\) \(i=\frac{2}{25}=0.08 \mathrm{~A}\) \(i=80 \mathrm{~mA}\)
COMEDK 2019
Semiconductor Electronics Material Devices and Simple Circuits
150765
Assertion: The value of current through \(p-n\) junction in the given figure will be \(10 \mathrm{~mA}\). Reason: In the above figure, \(p\)-side is at higher potential than n-side.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B The current, \(I=\frac{V}{R}=\frac{5-2}{300}=\frac{3}{300}=10^{-2} \mathrm{~A}\) \(I=10 \times 10^{-3}\) \(I=10 \mathrm{~mA} \text { (correct) }\) Reason-: \(\mathrm{P}\) side Voltage is \(5 \mathrm{~V}\) which is greater than \(\mathrm{N}\) side voltage is \(+2 \mathrm{~V}\) so correct.
AIIMS-2008
Semiconductor Electronics Material Devices and Simple Circuits
150766
A silicon diode has a threshold voltage of \(0.7 \mathrm{~V}\). If an input voltage given by \(2 \sin (\pi t)\) is supplied to a half wave rectifier circuit using this diode, the rectified output has a peak value.
1 \(2 \mathrm{~V}\)
2 \(14 . \mathrm{V}\)
3 \(1.3 \mathrm{~V}\)
4 \(0.7 \mathrm{~V}\)
Explanation:
C Given, Input voltage \(=2 \sin (\pi t)\) So, \(\quad\) Input peak voltage \(=2 \mathrm{~V}\) Barrier voltage \(=0.7 \mathrm{~V}\) \(\because\) The peak value of rectified output voltage \(=\) Peak voltage - threshold voltage \(=2-0.7\) \(=1.3 \mathrm{~V}\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150767
If current in diode is five times that in \(R_1\) and breakdown voltage of diode is 6 volt then find R.
1 \(\frac{1000}{3} \Omega\)
2 \(\frac{2000}{3} \Omega\)
3 \(\frac{4000}{5} \Omega\)
4 \(\frac{5000}{3} \Omega\)
Explanation:
B Current in diode is 5 times that in \(\mathrm{R}_1\) \(I_1=\frac{V}{R_1}=\frac{6}{1 \times 10^3} A\) \(I_1=6 \mathrm{~m} \mathrm{~A}\) Then, current in the diode \(\mathrm{I}_{\mathrm{D}}=6 \times 5=30 \mathrm{~mA}\) So, total current drawn from battery \(=6 \mathrm{~m} \mathrm{~A}+30 \mathrm{~m} \mathrm{~A} \Rightarrow 36 \mathrm{~m} \mathrm{~A}\) Now potential difference across resistance \(\mathrm{R}\) \(\mathrm{V}=24 \text { volt }\) So, \(\mathrm{V}=\mathrm{IR}\) \(24=36 \times 10^{-3} \mathrm{R}\) \(\mathrm{R}=\frac{2000}{3} \Omega\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150769
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is
1 \(0.75 \mathrm{~A}\)
2 \(0.5 \mathrm{~A}\)
3 \(0.25 \mathrm{~A}\)
4 zero
Explanation:
B In diode \(D_1\) the current pass through \(P\) side so current flow in diode \(D_1\) but in case of diode \(D_2\) there is no current flow because it is reverse bias. So \(\quad \mathrm{I}=\mathrm{V} / \mathrm{R}\) \(I=\frac{5}{10}\) \(I=0.5 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150764
The current I through \(\mathbf{1 0} \mathrm{ohm}\) resistor in the circuit given below is
1 \(50 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(80 \mathrm{~mA}\)
Explanation:
D In this circuit \(\mathrm{D}_2\) is reverse biased so, no current flow across \(\mathrm{D}_2\), now circuit become . Then, \(i=\frac{V}{R_{\text {eq }}}=\frac{2}{10+15}\) \(i=\frac{2}{25}=0.08 \mathrm{~A}\) \(i=80 \mathrm{~mA}\)
COMEDK 2019
Semiconductor Electronics Material Devices and Simple Circuits
150765
Assertion: The value of current through \(p-n\) junction in the given figure will be \(10 \mathrm{~mA}\). Reason: In the above figure, \(p\)-side is at higher potential than n-side.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B The current, \(I=\frac{V}{R}=\frac{5-2}{300}=\frac{3}{300}=10^{-2} \mathrm{~A}\) \(I=10 \times 10^{-3}\) \(I=10 \mathrm{~mA} \text { (correct) }\) Reason-: \(\mathrm{P}\) side Voltage is \(5 \mathrm{~V}\) which is greater than \(\mathrm{N}\) side voltage is \(+2 \mathrm{~V}\) so correct.
AIIMS-2008
Semiconductor Electronics Material Devices and Simple Circuits
150766
A silicon diode has a threshold voltage of \(0.7 \mathrm{~V}\). If an input voltage given by \(2 \sin (\pi t)\) is supplied to a half wave rectifier circuit using this diode, the rectified output has a peak value.
1 \(2 \mathrm{~V}\)
2 \(14 . \mathrm{V}\)
3 \(1.3 \mathrm{~V}\)
4 \(0.7 \mathrm{~V}\)
Explanation:
C Given, Input voltage \(=2 \sin (\pi t)\) So, \(\quad\) Input peak voltage \(=2 \mathrm{~V}\) Barrier voltage \(=0.7 \mathrm{~V}\) \(\because\) The peak value of rectified output voltage \(=\) Peak voltage - threshold voltage \(=2-0.7\) \(=1.3 \mathrm{~V}\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150767
If current in diode is five times that in \(R_1\) and breakdown voltage of diode is 6 volt then find R.
1 \(\frac{1000}{3} \Omega\)
2 \(\frac{2000}{3} \Omega\)
3 \(\frac{4000}{5} \Omega\)
4 \(\frac{5000}{3} \Omega\)
Explanation:
B Current in diode is 5 times that in \(\mathrm{R}_1\) \(I_1=\frac{V}{R_1}=\frac{6}{1 \times 10^3} A\) \(I_1=6 \mathrm{~m} \mathrm{~A}\) Then, current in the diode \(\mathrm{I}_{\mathrm{D}}=6 \times 5=30 \mathrm{~mA}\) So, total current drawn from battery \(=6 \mathrm{~m} \mathrm{~A}+30 \mathrm{~m} \mathrm{~A} \Rightarrow 36 \mathrm{~m} \mathrm{~A}\) Now potential difference across resistance \(\mathrm{R}\) \(\mathrm{V}=24 \text { volt }\) So, \(\mathrm{V}=\mathrm{IR}\) \(24=36 \times 10^{-3} \mathrm{R}\) \(\mathrm{R}=\frac{2000}{3} \Omega\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150769
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is
1 \(0.75 \mathrm{~A}\)
2 \(0.5 \mathrm{~A}\)
3 \(0.25 \mathrm{~A}\)
4 zero
Explanation:
B In diode \(D_1\) the current pass through \(P\) side so current flow in diode \(D_1\) but in case of diode \(D_2\) there is no current flow because it is reverse bias. So \(\quad \mathrm{I}=\mathrm{V} / \mathrm{R}\) \(I=\frac{5}{10}\) \(I=0.5 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150764
The current I through \(\mathbf{1 0} \mathrm{ohm}\) resistor in the circuit given below is
1 \(50 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(80 \mathrm{~mA}\)
Explanation:
D In this circuit \(\mathrm{D}_2\) is reverse biased so, no current flow across \(\mathrm{D}_2\), now circuit become . Then, \(i=\frac{V}{R_{\text {eq }}}=\frac{2}{10+15}\) \(i=\frac{2}{25}=0.08 \mathrm{~A}\) \(i=80 \mathrm{~mA}\)
COMEDK 2019
Semiconductor Electronics Material Devices and Simple Circuits
150765
Assertion: The value of current through \(p-n\) junction in the given figure will be \(10 \mathrm{~mA}\). Reason: In the above figure, \(p\)-side is at higher potential than n-side.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B The current, \(I=\frac{V}{R}=\frac{5-2}{300}=\frac{3}{300}=10^{-2} \mathrm{~A}\) \(I=10 \times 10^{-3}\) \(I=10 \mathrm{~mA} \text { (correct) }\) Reason-: \(\mathrm{P}\) side Voltage is \(5 \mathrm{~V}\) which is greater than \(\mathrm{N}\) side voltage is \(+2 \mathrm{~V}\) so correct.
AIIMS-2008
Semiconductor Electronics Material Devices and Simple Circuits
150766
A silicon diode has a threshold voltage of \(0.7 \mathrm{~V}\). If an input voltage given by \(2 \sin (\pi t)\) is supplied to a half wave rectifier circuit using this diode, the rectified output has a peak value.
1 \(2 \mathrm{~V}\)
2 \(14 . \mathrm{V}\)
3 \(1.3 \mathrm{~V}\)
4 \(0.7 \mathrm{~V}\)
Explanation:
C Given, Input voltage \(=2 \sin (\pi t)\) So, \(\quad\) Input peak voltage \(=2 \mathrm{~V}\) Barrier voltage \(=0.7 \mathrm{~V}\) \(\because\) The peak value of rectified output voltage \(=\) Peak voltage - threshold voltage \(=2-0.7\) \(=1.3 \mathrm{~V}\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150767
If current in diode is five times that in \(R_1\) and breakdown voltage of diode is 6 volt then find R.
1 \(\frac{1000}{3} \Omega\)
2 \(\frac{2000}{3} \Omega\)
3 \(\frac{4000}{5} \Omega\)
4 \(\frac{5000}{3} \Omega\)
Explanation:
B Current in diode is 5 times that in \(\mathrm{R}_1\) \(I_1=\frac{V}{R_1}=\frac{6}{1 \times 10^3} A\) \(I_1=6 \mathrm{~m} \mathrm{~A}\) Then, current in the diode \(\mathrm{I}_{\mathrm{D}}=6 \times 5=30 \mathrm{~mA}\) So, total current drawn from battery \(=6 \mathrm{~m} \mathrm{~A}+30 \mathrm{~m} \mathrm{~A} \Rightarrow 36 \mathrm{~m} \mathrm{~A}\) Now potential difference across resistance \(\mathrm{R}\) \(\mathrm{V}=24 \text { volt }\) So, \(\mathrm{V}=\mathrm{IR}\) \(24=36 \times 10^{-3} \mathrm{R}\) \(\mathrm{R}=\frac{2000}{3} \Omega\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150769
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is
1 \(0.75 \mathrm{~A}\)
2 \(0.5 \mathrm{~A}\)
3 \(0.25 \mathrm{~A}\)
4 zero
Explanation:
B In diode \(D_1\) the current pass through \(P\) side so current flow in diode \(D_1\) but in case of diode \(D_2\) there is no current flow because it is reverse bias. So \(\quad \mathrm{I}=\mathrm{V} / \mathrm{R}\) \(I=\frac{5}{10}\) \(I=0.5 \mathrm{~A}\)
