Semiconductor Electronics Material Devices and Simple Circuits
150744
A full-wave rectifier circuit with an \(\mathrm{AC}\) input is shown The output voltage across \(R_L\) is represented as
1
2
3
4
5
Explanation:
E Given here full wave rectifier. In full wave rectifier minimum two diodes are connected in circuit. In which one diode acts on positive of cycle and second diode acts on negative half cycle So, we get continuous wave. Then the output will be,
Kerala CEE - 2010
Semiconductor Electronics Material Devices and Simple Circuits
150745
In the diagram, the input \(\mathrm{AC}\) is across the terminals \(A\) and \(C\). The output across B and D is
1 same as the input
2 half wave rectified
3 zero
4 full wave rectified
5 steady DC
Explanation:
D 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 \(\rightarrow D_1 \& D_3\) 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 \(10 \Omega\) resistor in the circuit given below is
1 \(50 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(80 \mathrm{~mA}\)
5 \(35 \mathrm{~mA}\)
Explanation:
D Here \(\mathrm{D}_2\) is reversed biased while \(\mathrm{D}_1\) is forward biased so current flow only \(D_1\) Then, \(I=\frac{V}{R}=\frac{2}{10+15}=\frac{2}{25}=0.08 \mathrm{~A}\) \(\mathrm{I}=80 \mathrm{~mA}\)
Kerala CEE - 2008
Semiconductor Electronics Material Devices and Simple Circuits
150747
When the forward bias voltage of a diode is changed from \(0.6 \mathrm{~V}\) to \(0.7 \mathrm{~V}\), the current changes from \(5 \mathrm{~mA}\) to \(15 \mathrm{~mA}\). Then its forward bias resistance is
1 \(0.01 \Omega\)
2 \(0.1 \Omega\)
3 \(10 \Omega\)
4 \(100 \Omega\)
5 \(0.2 \Omega\)
Explanation:
C Given that, \(\mathrm{V}_1=0.6 \mathrm{~V} \mathrm{~V}_2=0.7 \mathrm{~V}\) \(\mathrm{I}_1=5 \mathrm{~mA} \mathrm{I}_2=15 \mathrm{~mA}\) Then forward biased resistance will be, \(\mathrm{r}=\frac{\mathrm{dV}}{\mathrm{dI}} =\frac{0.7-0.6}{15 \mathrm{~mA}-5 \mathrm{~mA}}=\frac{0.1}{10 \times 10^{-3} \mathrm{~A}}\) \(\mathrm{r} =0.1 \times 10^2 \Omega\) \(\mathrm{r} =10 \Omega\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Semiconductor Electronics Material Devices and Simple Circuits
150744
A full-wave rectifier circuit with an \(\mathrm{AC}\) input is shown The output voltage across \(R_L\) is represented as
1
2
3
4
5
Explanation:
E Given here full wave rectifier. In full wave rectifier minimum two diodes are connected in circuit. In which one diode acts on positive of cycle and second diode acts on negative half cycle So, we get continuous wave. Then the output will be,
Kerala CEE - 2010
Semiconductor Electronics Material Devices and Simple Circuits
150745
In the diagram, the input \(\mathrm{AC}\) is across the terminals \(A\) and \(C\). The output across B and D is
1 same as the input
2 half wave rectified
3 zero
4 full wave rectified
5 steady DC
Explanation:
D 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 \(\rightarrow D_1 \& D_3\) 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 \(10 \Omega\) resistor in the circuit given below is
1 \(50 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(80 \mathrm{~mA}\)
5 \(35 \mathrm{~mA}\)
Explanation:
D Here \(\mathrm{D}_2\) is reversed biased while \(\mathrm{D}_1\) is forward biased so current flow only \(D_1\) Then, \(I=\frac{V}{R}=\frac{2}{10+15}=\frac{2}{25}=0.08 \mathrm{~A}\) \(\mathrm{I}=80 \mathrm{~mA}\)
Kerala CEE - 2008
Semiconductor Electronics Material Devices and Simple Circuits
150747
When the forward bias voltage of a diode is changed from \(0.6 \mathrm{~V}\) to \(0.7 \mathrm{~V}\), the current changes from \(5 \mathrm{~mA}\) to \(15 \mathrm{~mA}\). Then its forward bias resistance is
1 \(0.01 \Omega\)
2 \(0.1 \Omega\)
3 \(10 \Omega\)
4 \(100 \Omega\)
5 \(0.2 \Omega\)
Explanation:
C Given that, \(\mathrm{V}_1=0.6 \mathrm{~V} \mathrm{~V}_2=0.7 \mathrm{~V}\) \(\mathrm{I}_1=5 \mathrm{~mA} \mathrm{I}_2=15 \mathrm{~mA}\) Then forward biased resistance will be, \(\mathrm{r}=\frac{\mathrm{dV}}{\mathrm{dI}} =\frac{0.7-0.6}{15 \mathrm{~mA}-5 \mathrm{~mA}}=\frac{0.1}{10 \times 10^{-3} \mathrm{~A}}\) \(\mathrm{r} =0.1 \times 10^2 \Omega\) \(\mathrm{r} =10 \Omega\)
Semiconductor Electronics Material Devices and Simple Circuits
150744
A full-wave rectifier circuit with an \(\mathrm{AC}\) input is shown The output voltage across \(R_L\) is represented as
1
2
3
4
5
Explanation:
E Given here full wave rectifier. In full wave rectifier minimum two diodes are connected in circuit. In which one diode acts on positive of cycle and second diode acts on negative half cycle So, we get continuous wave. Then the output will be,
Kerala CEE - 2010
Semiconductor Electronics Material Devices and Simple Circuits
150745
In the diagram, the input \(\mathrm{AC}\) is across the terminals \(A\) and \(C\). The output across B and D is
1 same as the input
2 half wave rectified
3 zero
4 full wave rectified
5 steady DC
Explanation:
D 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 \(\rightarrow D_1 \& D_3\) 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 \(10 \Omega\) resistor in the circuit given below is
1 \(50 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(80 \mathrm{~mA}\)
5 \(35 \mathrm{~mA}\)
Explanation:
D Here \(\mathrm{D}_2\) is reversed biased while \(\mathrm{D}_1\) is forward biased so current flow only \(D_1\) Then, \(I=\frac{V}{R}=\frac{2}{10+15}=\frac{2}{25}=0.08 \mathrm{~A}\) \(\mathrm{I}=80 \mathrm{~mA}\)
Kerala CEE - 2008
Semiconductor Electronics Material Devices and Simple Circuits
150747
When the forward bias voltage of a diode is changed from \(0.6 \mathrm{~V}\) to \(0.7 \mathrm{~V}\), the current changes from \(5 \mathrm{~mA}\) to \(15 \mathrm{~mA}\). Then its forward bias resistance is
1 \(0.01 \Omega\)
2 \(0.1 \Omega\)
3 \(10 \Omega\)
4 \(100 \Omega\)
5 \(0.2 \Omega\)
Explanation:
C Given that, \(\mathrm{V}_1=0.6 \mathrm{~V} \mathrm{~V}_2=0.7 \mathrm{~V}\) \(\mathrm{I}_1=5 \mathrm{~mA} \mathrm{I}_2=15 \mathrm{~mA}\) Then forward biased resistance will be, \(\mathrm{r}=\frac{\mathrm{dV}}{\mathrm{dI}} =\frac{0.7-0.6}{15 \mathrm{~mA}-5 \mathrm{~mA}}=\frac{0.1}{10 \times 10^{-3} \mathrm{~A}}\) \(\mathrm{r} =0.1 \times 10^2 \Omega\) \(\mathrm{r} =10 \Omega\)
Semiconductor Electronics Material Devices and Simple Circuits
150744
A full-wave rectifier circuit with an \(\mathrm{AC}\) input is shown The output voltage across \(R_L\) is represented as
1
2
3
4
5
Explanation:
E Given here full wave rectifier. In full wave rectifier minimum two diodes are connected in circuit. In which one diode acts on positive of cycle and second diode acts on negative half cycle So, we get continuous wave. Then the output will be,
Kerala CEE - 2010
Semiconductor Electronics Material Devices and Simple Circuits
150745
In the diagram, the input \(\mathrm{AC}\) is across the terminals \(A\) and \(C\). The output across B and D is
1 same as the input
2 half wave rectified
3 zero
4 full wave rectified
5 steady DC
Explanation:
D 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 \(\rightarrow D_1 \& D_3\) 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 \(10 \Omega\) resistor in the circuit given below is
1 \(50 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(80 \mathrm{~mA}\)
5 \(35 \mathrm{~mA}\)
Explanation:
D Here \(\mathrm{D}_2\) is reversed biased while \(\mathrm{D}_1\) is forward biased so current flow only \(D_1\) Then, \(I=\frac{V}{R}=\frac{2}{10+15}=\frac{2}{25}=0.08 \mathrm{~A}\) \(\mathrm{I}=80 \mathrm{~mA}\)
Kerala CEE - 2008
Semiconductor Electronics Material Devices and Simple Circuits
150747
When the forward bias voltage of a diode is changed from \(0.6 \mathrm{~V}\) to \(0.7 \mathrm{~V}\), the current changes from \(5 \mathrm{~mA}\) to \(15 \mathrm{~mA}\). Then its forward bias resistance is
1 \(0.01 \Omega\)
2 \(0.1 \Omega\)
3 \(10 \Omega\)
4 \(100 \Omega\)
5 \(0.2 \Omega\)
Explanation:
C Given that, \(\mathrm{V}_1=0.6 \mathrm{~V} \mathrm{~V}_2=0.7 \mathrm{~V}\) \(\mathrm{I}_1=5 \mathrm{~mA} \mathrm{I}_2=15 \mathrm{~mA}\) Then forward biased resistance will be, \(\mathrm{r}=\frac{\mathrm{dV}}{\mathrm{dI}} =\frac{0.7-0.6}{15 \mathrm{~mA}-5 \mathrm{~mA}}=\frac{0.1}{10 \times 10^{-3} \mathrm{~A}}\) \(\mathrm{r} =0.1 \times 10^2 \Omega\) \(\mathrm{r} =10 \Omega\)
