PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365341
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery in circuit
1 \(0.5\,A\)
2 \(0.25\,A\)
3 \(1\,A\)
4 \(2\,A\)
Explanation:
Diode \({D_{1}}\) and \({D_{3}}\) are forward biased and \({D_{2}}\) is reverse biased. So, circuit will look like as shown below We have, \(i=\dfrac{5}{10+10} \Rightarrow i=\dfrac{5}{20}=0.25 {~A}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365342
If each diode has a forward bias resistance of \(25 \Omega\) in the below circuit, which of the following option is correct?
1 \(\dfrac{I_{1}}{I_{2}}=1\)
2 \(\dfrac{I_{2}}{I_{3}}=1\)
3 \(\dfrac{I_{1}}{I_{2}}=2\)
4 \(\dfrac{I_{3}}{I_{4}}=1\)
Explanation:
Diode \(D_{2}\) is reversed biased. So, the new circuit is given by For the circuit, two resistors of resistance \(125 \Omega\) are in parallel. Resultant of \(125 \Omega\) is in series with \(25 \Omega\). \(R_{e q}=\dfrac{150 \times 150}{300}+25=100 \Omega\) \({I_1} = \frac{5}{{100}} = 0.05\;A\) \({I_2} = {I_4} = \frac{{0.05}}{2} = 0.025\;A\) \(\dfrac{I_{1}}{I_{2}}=2\)
JEE - 2023
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365343
The resistance of a germanium junction diode whose \(V-I\) is shown in figure is\(\left( {{V_k} = 0.3\;V} \right)\)
From graph, Resistance of the germanium junction diode is \(R = \frac{{\Delta V}}{{\Delta I}} = \frac{{2.3\;V - 0.3\;V}}{{10\;mA - 0}} = \frac{{2\;V}}{{10 \times {{10}^{ - 3}}\;A}}\) \(=0.2 \times 10^{3} \Omega=0.2\, \mathrm{k} \Omega\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365344
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
1 \(2.0\,A\)
2 \(1.33\,A\)
3 \(1.71\,A\)
4 \(2.31\,A\)
Explanation:
In the given circuit diode \({D_1}\) is forward biased and \({D_2}\) is reverse biased. \(\therefore \) The effictive resistance of \({D_1}\) is zero and that of \({D_2}\) is infinite as the diodes \({D_1}\) and \({D_2}\) are ideal. No current flows through \({D_2}\) and the only active resistances in the circuit are the \(3\Omega \) and \(4\Omega \) resisances as the reverse biased \({D_2}\) effictively disconnects the \(2\Omega \) resistance. \(\therefore \) The current in the circuit is \(I = \frac{{12V}}{{4\Omega + 3\Omega }} = \frac{{12V}}{{7\Omega }} = 1.71A\)
KCET - 2015
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365345
When a \(p-n\) junction diode is in forward bias, which type of charge carriers flows in the connecting wire?
1 Protons
2 Holes
3 Free electrons
4 Ions
Explanation:
In forward bias of \(p-n\) junction diode electrons are attracted towards positive terminal of battery, electrons get captured into the hole and leaves the hole. Correct option is (3).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365341
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery in circuit
1 \(0.5\,A\)
2 \(0.25\,A\)
3 \(1\,A\)
4 \(2\,A\)
Explanation:
Diode \({D_{1}}\) and \({D_{3}}\) are forward biased and \({D_{2}}\) is reverse biased. So, circuit will look like as shown below We have, \(i=\dfrac{5}{10+10} \Rightarrow i=\dfrac{5}{20}=0.25 {~A}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365342
If each diode has a forward bias resistance of \(25 \Omega\) in the below circuit, which of the following option is correct?
1 \(\dfrac{I_{1}}{I_{2}}=1\)
2 \(\dfrac{I_{2}}{I_{3}}=1\)
3 \(\dfrac{I_{1}}{I_{2}}=2\)
4 \(\dfrac{I_{3}}{I_{4}}=1\)
Explanation:
Diode \(D_{2}\) is reversed biased. So, the new circuit is given by For the circuit, two resistors of resistance \(125 \Omega\) are in parallel. Resultant of \(125 \Omega\) is in series with \(25 \Omega\). \(R_{e q}=\dfrac{150 \times 150}{300}+25=100 \Omega\) \({I_1} = \frac{5}{{100}} = 0.05\;A\) \({I_2} = {I_4} = \frac{{0.05}}{2} = 0.025\;A\) \(\dfrac{I_{1}}{I_{2}}=2\)
JEE - 2023
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365343
The resistance of a germanium junction diode whose \(V-I\) is shown in figure is\(\left( {{V_k} = 0.3\;V} \right)\)
From graph, Resistance of the germanium junction diode is \(R = \frac{{\Delta V}}{{\Delta I}} = \frac{{2.3\;V - 0.3\;V}}{{10\;mA - 0}} = \frac{{2\;V}}{{10 \times {{10}^{ - 3}}\;A}}\) \(=0.2 \times 10^{3} \Omega=0.2\, \mathrm{k} \Omega\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365344
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
1 \(2.0\,A\)
2 \(1.33\,A\)
3 \(1.71\,A\)
4 \(2.31\,A\)
Explanation:
In the given circuit diode \({D_1}\) is forward biased and \({D_2}\) is reverse biased. \(\therefore \) The effictive resistance of \({D_1}\) is zero and that of \({D_2}\) is infinite as the diodes \({D_1}\) and \({D_2}\) are ideal. No current flows through \({D_2}\) and the only active resistances in the circuit are the \(3\Omega \) and \(4\Omega \) resisances as the reverse biased \({D_2}\) effictively disconnects the \(2\Omega \) resistance. \(\therefore \) The current in the circuit is \(I = \frac{{12V}}{{4\Omega + 3\Omega }} = \frac{{12V}}{{7\Omega }} = 1.71A\)
KCET - 2015
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365345
When a \(p-n\) junction diode is in forward bias, which type of charge carriers flows in the connecting wire?
1 Protons
2 Holes
3 Free electrons
4 Ions
Explanation:
In forward bias of \(p-n\) junction diode electrons are attracted towards positive terminal of battery, electrons get captured into the hole and leaves the hole. Correct option is (3).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365341
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery in circuit
1 \(0.5\,A\)
2 \(0.25\,A\)
3 \(1\,A\)
4 \(2\,A\)
Explanation:
Diode \({D_{1}}\) and \({D_{3}}\) are forward biased and \({D_{2}}\) is reverse biased. So, circuit will look like as shown below We have, \(i=\dfrac{5}{10+10} \Rightarrow i=\dfrac{5}{20}=0.25 {~A}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365342
If each diode has a forward bias resistance of \(25 \Omega\) in the below circuit, which of the following option is correct?
1 \(\dfrac{I_{1}}{I_{2}}=1\)
2 \(\dfrac{I_{2}}{I_{3}}=1\)
3 \(\dfrac{I_{1}}{I_{2}}=2\)
4 \(\dfrac{I_{3}}{I_{4}}=1\)
Explanation:
Diode \(D_{2}\) is reversed biased. So, the new circuit is given by For the circuit, two resistors of resistance \(125 \Omega\) are in parallel. Resultant of \(125 \Omega\) is in series with \(25 \Omega\). \(R_{e q}=\dfrac{150 \times 150}{300}+25=100 \Omega\) \({I_1} = \frac{5}{{100}} = 0.05\;A\) \({I_2} = {I_4} = \frac{{0.05}}{2} = 0.025\;A\) \(\dfrac{I_{1}}{I_{2}}=2\)
JEE - 2023
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365343
The resistance of a germanium junction diode whose \(V-I\) is shown in figure is\(\left( {{V_k} = 0.3\;V} \right)\)
From graph, Resistance of the germanium junction diode is \(R = \frac{{\Delta V}}{{\Delta I}} = \frac{{2.3\;V - 0.3\;V}}{{10\;mA - 0}} = \frac{{2\;V}}{{10 \times {{10}^{ - 3}}\;A}}\) \(=0.2 \times 10^{3} \Omega=0.2\, \mathrm{k} \Omega\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365344
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
1 \(2.0\,A\)
2 \(1.33\,A\)
3 \(1.71\,A\)
4 \(2.31\,A\)
Explanation:
In the given circuit diode \({D_1}\) is forward biased and \({D_2}\) is reverse biased. \(\therefore \) The effictive resistance of \({D_1}\) is zero and that of \({D_2}\) is infinite as the diodes \({D_1}\) and \({D_2}\) are ideal. No current flows through \({D_2}\) and the only active resistances in the circuit are the \(3\Omega \) and \(4\Omega \) resisances as the reverse biased \({D_2}\) effictively disconnects the \(2\Omega \) resistance. \(\therefore \) The current in the circuit is \(I = \frac{{12V}}{{4\Omega + 3\Omega }} = \frac{{12V}}{{7\Omega }} = 1.71A\)
KCET - 2015
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365345
When a \(p-n\) junction diode is in forward bias, which type of charge carriers flows in the connecting wire?
1 Protons
2 Holes
3 Free electrons
4 Ions
Explanation:
In forward bias of \(p-n\) junction diode electrons are attracted towards positive terminal of battery, electrons get captured into the hole and leaves the hole. Correct option is (3).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365341
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery in circuit
1 \(0.5\,A\)
2 \(0.25\,A\)
3 \(1\,A\)
4 \(2\,A\)
Explanation:
Diode \({D_{1}}\) and \({D_{3}}\) are forward biased and \({D_{2}}\) is reverse biased. So, circuit will look like as shown below We have, \(i=\dfrac{5}{10+10} \Rightarrow i=\dfrac{5}{20}=0.25 {~A}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365342
If each diode has a forward bias resistance of \(25 \Omega\) in the below circuit, which of the following option is correct?
1 \(\dfrac{I_{1}}{I_{2}}=1\)
2 \(\dfrac{I_{2}}{I_{3}}=1\)
3 \(\dfrac{I_{1}}{I_{2}}=2\)
4 \(\dfrac{I_{3}}{I_{4}}=1\)
Explanation:
Diode \(D_{2}\) is reversed biased. So, the new circuit is given by For the circuit, two resistors of resistance \(125 \Omega\) are in parallel. Resultant of \(125 \Omega\) is in series with \(25 \Omega\). \(R_{e q}=\dfrac{150 \times 150}{300}+25=100 \Omega\) \({I_1} = \frac{5}{{100}} = 0.05\;A\) \({I_2} = {I_4} = \frac{{0.05}}{2} = 0.025\;A\) \(\dfrac{I_{1}}{I_{2}}=2\)
JEE - 2023
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365343
The resistance of a germanium junction diode whose \(V-I\) is shown in figure is\(\left( {{V_k} = 0.3\;V} \right)\)
From graph, Resistance of the germanium junction diode is \(R = \frac{{\Delta V}}{{\Delta I}} = \frac{{2.3\;V - 0.3\;V}}{{10\;mA - 0}} = \frac{{2\;V}}{{10 \times {{10}^{ - 3}}\;A}}\) \(=0.2 \times 10^{3} \Omega=0.2\, \mathrm{k} \Omega\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365344
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
1 \(2.0\,A\)
2 \(1.33\,A\)
3 \(1.71\,A\)
4 \(2.31\,A\)
Explanation:
In the given circuit diode \({D_1}\) is forward biased and \({D_2}\) is reverse biased. \(\therefore \) The effictive resistance of \({D_1}\) is zero and that of \({D_2}\) is infinite as the diodes \({D_1}\) and \({D_2}\) are ideal. No current flows through \({D_2}\) and the only active resistances in the circuit are the \(3\Omega \) and \(4\Omega \) resisances as the reverse biased \({D_2}\) effictively disconnects the \(2\Omega \) resistance. \(\therefore \) The current in the circuit is \(I = \frac{{12V}}{{4\Omega + 3\Omega }} = \frac{{12V}}{{7\Omega }} = 1.71A\)
KCET - 2015
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365345
When a \(p-n\) junction diode is in forward bias, which type of charge carriers flows in the connecting wire?
1 Protons
2 Holes
3 Free electrons
4 Ions
Explanation:
In forward bias of \(p-n\) junction diode electrons are attracted towards positive terminal of battery, electrons get captured into the hole and leaves the hole. Correct option is (3).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365341
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery in circuit
1 \(0.5\,A\)
2 \(0.25\,A\)
3 \(1\,A\)
4 \(2\,A\)
Explanation:
Diode \({D_{1}}\) and \({D_{3}}\) are forward biased and \({D_{2}}\) is reverse biased. So, circuit will look like as shown below We have, \(i=\dfrac{5}{10+10} \Rightarrow i=\dfrac{5}{20}=0.25 {~A}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365342
If each diode has a forward bias resistance of \(25 \Omega\) in the below circuit, which of the following option is correct?
1 \(\dfrac{I_{1}}{I_{2}}=1\)
2 \(\dfrac{I_{2}}{I_{3}}=1\)
3 \(\dfrac{I_{1}}{I_{2}}=2\)
4 \(\dfrac{I_{3}}{I_{4}}=1\)
Explanation:
Diode \(D_{2}\) is reversed biased. So, the new circuit is given by For the circuit, two resistors of resistance \(125 \Omega\) are in parallel. Resultant of \(125 \Omega\) is in series with \(25 \Omega\). \(R_{e q}=\dfrac{150 \times 150}{300}+25=100 \Omega\) \({I_1} = \frac{5}{{100}} = 0.05\;A\) \({I_2} = {I_4} = \frac{{0.05}}{2} = 0.025\;A\) \(\dfrac{I_{1}}{I_{2}}=2\)
JEE - 2023
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365343
The resistance of a germanium junction diode whose \(V-I\) is shown in figure is\(\left( {{V_k} = 0.3\;V} \right)\)
From graph, Resistance of the germanium junction diode is \(R = \frac{{\Delta V}}{{\Delta I}} = \frac{{2.3\;V - 0.3\;V}}{{10\;mA - 0}} = \frac{{2\;V}}{{10 \times {{10}^{ - 3}}\;A}}\) \(=0.2 \times 10^{3} \Omega=0.2\, \mathrm{k} \Omega\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365344
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
1 \(2.0\,A\)
2 \(1.33\,A\)
3 \(1.71\,A\)
4 \(2.31\,A\)
Explanation:
In the given circuit diode \({D_1}\) is forward biased and \({D_2}\) is reverse biased. \(\therefore \) The effictive resistance of \({D_1}\) is zero and that of \({D_2}\) is infinite as the diodes \({D_1}\) and \({D_2}\) are ideal. No current flows through \({D_2}\) and the only active resistances in the circuit are the \(3\Omega \) and \(4\Omega \) resisances as the reverse biased \({D_2}\) effictively disconnects the \(2\Omega \) resistance. \(\therefore \) The current in the circuit is \(I = \frac{{12V}}{{4\Omega + 3\Omega }} = \frac{{12V}}{{7\Omega }} = 1.71A\)
KCET - 2015
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365345
When a \(p-n\) junction diode is in forward bias, which type of charge carriers flows in the connecting wire?
1 Protons
2 Holes
3 Free electrons
4 Ions
Explanation:
In forward bias of \(p-n\) junction diode electrons are attracted towards positive terminal of battery, electrons get captured into the hole and leaves the hole. Correct option is (3).
