Semiconductor Electronics Material Devices and Simple Circuits
150749
For the given circuit shown below, to act as full wave rectifier, the \(\mathrm{AC}\) input should be connected across and and the DC. output would appear across and
1 B and D and A and C
2 B and A and C and D
3 C and A and B and D
4 C and D and B and A
5 none of the above
Explanation:
B Let naming the diodes \(\mathrm{D}_1, \mathrm{D}_2, \mathrm{D}_3\) and \(\mathrm{D}_4\) According to the question circuit is act as full wave rectifier. In above figure, when we apply \(\mathrm{AC}\) input across \(\mathrm{BD}\). For positive half cycle diode \(D_1 \& D_3\) act as forward bias and \(\mathrm{D}_2 \& \mathrm{D}_4\) act as reverse bias. For negative half cycle \(D_1 \& D_3\) act as a reverse bias and \(D_2 \& D_4\) act as forward bias. Hence, AC input should be connected across B and D and the DC. Output would appear across A and C.
Kerala CEE 2004
Semiconductor Electronics Material Devices and Simple Circuits
150750
In the circuit of figure, treat diode as ideal, current in the \(4 \Omega\) resistor is
1 \(2 \mathrm{~A}\)
2 \(3 \mathrm{~A}\)
3 \(\frac{12}{7} \mathrm{~A}\)
4 \(\frac{30}{13} \mathrm{~A}\)
Explanation:
B From above figure, we can see diode \(\mathrm{D}_1\) is forward bias so, current flow through it and \(\mathrm{D}_2\) is reverse bias and it acts like a open circuit. Then resistance \(2 \Omega\) and \(4 \Omega\) are in series connection. \(\therefore \mathrm{R}_{\mathrm{eq}}=2+4=6 \Omega\) Now, current in \(4 \Omega\) resistance is \(I=\frac{V}{R_{e q}}\) \(I=\frac{12}{6}=2 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150752
A diode is connected to \(220 \mathrm{~V}\) (rms) \(\mathrm{AC}\) in series with a capacitor as shown in figure. The voltage across the capacitor is
1 \(220 \mathrm{~V}\)
2 \(110 \mathrm{~V}\)
3 \(311.1 \mathrm{~V}\)
4 \(\frac{220}{\sqrt{2}} \mathrm{~V}\)
Explanation:
D In this circuit diode D conduct in positive half cycle and in negative half cycle this is open circuit so it behaves as half wave rectifier. The voltage across the capacitor, \(\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}_0}{2}=\frac{220 \sqrt{2}}{2} \Rightarrow \frac{220}{\sqrt{2}}\)
JCECE-2012
Semiconductor Electronics Material Devices and Simple Circuits
150761
In a transistor connected in a common emitter \(\operatorname{mode}, R_C=4 \mathrm{k} \Omega, R_1=1 \mathrm{k} \Omega, I_C=1 \mathrm{~mA}\) and \(I_b=\) \(20 \mu \mathrm{A}\). The voltage gain is
1 100
2 200
3 300
4 400
Explanation:
B Given that, \(\mathrm{R}_{\mathrm{c}}=4 \times 10^3 \Omega \mathrm{R}_1=1 \times 10^3 \Omega\) \(\mathrm{I}_{\mathrm{C}}=1 \times 10^{-3} \mathrm{I}_{\mathrm{B}}=20 \times 10^{-6} \mathrm{~A}\) We know that, voltage gain, \(A_V=\frac{\beta \mathrm{R}_C}{R_1}=\frac{I_C}{I_B} \frac{R_c}{R_1} \quad\left(\because \beta=\frac{I_C}{I_B}\right)\) \(A_v=\frac{1 \times 10^{-3} \times 4 \times 10^3}{20 \times 10^{-6} \times 1 \times 10^3}\) \(A_V=200\)
Semiconductor Electronics Material Devices and Simple Circuits
150749
For the given circuit shown below, to act as full wave rectifier, the \(\mathrm{AC}\) input should be connected across and and the DC. output would appear across and
1 B and D and A and C
2 B and A and C and D
3 C and A and B and D
4 C and D and B and A
5 none of the above
Explanation:
B Let naming the diodes \(\mathrm{D}_1, \mathrm{D}_2, \mathrm{D}_3\) and \(\mathrm{D}_4\) According to the question circuit is act as full wave rectifier. In above figure, when we apply \(\mathrm{AC}\) input across \(\mathrm{BD}\). For positive half cycle diode \(D_1 \& D_3\) act as forward bias and \(\mathrm{D}_2 \& \mathrm{D}_4\) act as reverse bias. For negative half cycle \(D_1 \& D_3\) act as a reverse bias and \(D_2 \& D_4\) act as forward bias. Hence, AC input should be connected across B and D and the DC. Output would appear across A and C.
Kerala CEE 2004
Semiconductor Electronics Material Devices and Simple Circuits
150750
In the circuit of figure, treat diode as ideal, current in the \(4 \Omega\) resistor is
1 \(2 \mathrm{~A}\)
2 \(3 \mathrm{~A}\)
3 \(\frac{12}{7} \mathrm{~A}\)
4 \(\frac{30}{13} \mathrm{~A}\)
Explanation:
B From above figure, we can see diode \(\mathrm{D}_1\) is forward bias so, current flow through it and \(\mathrm{D}_2\) is reverse bias and it acts like a open circuit. Then resistance \(2 \Omega\) and \(4 \Omega\) are in series connection. \(\therefore \mathrm{R}_{\mathrm{eq}}=2+4=6 \Omega\) Now, current in \(4 \Omega\) resistance is \(I=\frac{V}{R_{e q}}\) \(I=\frac{12}{6}=2 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150752
A diode is connected to \(220 \mathrm{~V}\) (rms) \(\mathrm{AC}\) in series with a capacitor as shown in figure. The voltage across the capacitor is
1 \(220 \mathrm{~V}\)
2 \(110 \mathrm{~V}\)
3 \(311.1 \mathrm{~V}\)
4 \(\frac{220}{\sqrt{2}} \mathrm{~V}\)
Explanation:
D In this circuit diode D conduct in positive half cycle and in negative half cycle this is open circuit so it behaves as half wave rectifier. The voltage across the capacitor, \(\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}_0}{2}=\frac{220 \sqrt{2}}{2} \Rightarrow \frac{220}{\sqrt{2}}\)
JCECE-2012
Semiconductor Electronics Material Devices and Simple Circuits
150761
In a transistor connected in a common emitter \(\operatorname{mode}, R_C=4 \mathrm{k} \Omega, R_1=1 \mathrm{k} \Omega, I_C=1 \mathrm{~mA}\) and \(I_b=\) \(20 \mu \mathrm{A}\). The voltage gain is
1 100
2 200
3 300
4 400
Explanation:
B Given that, \(\mathrm{R}_{\mathrm{c}}=4 \times 10^3 \Omega \mathrm{R}_1=1 \times 10^3 \Omega\) \(\mathrm{I}_{\mathrm{C}}=1 \times 10^{-3} \mathrm{I}_{\mathrm{B}}=20 \times 10^{-6} \mathrm{~A}\) We know that, voltage gain, \(A_V=\frac{\beta \mathrm{R}_C}{R_1}=\frac{I_C}{I_B} \frac{R_c}{R_1} \quad\left(\because \beta=\frac{I_C}{I_B}\right)\) \(A_v=\frac{1 \times 10^{-3} \times 4 \times 10^3}{20 \times 10^{-6} \times 1 \times 10^3}\) \(A_V=200\)
Semiconductor Electronics Material Devices and Simple Circuits
150749
For the given circuit shown below, to act as full wave rectifier, the \(\mathrm{AC}\) input should be connected across and and the DC. output would appear across and
1 B and D and A and C
2 B and A and C and D
3 C and A and B and D
4 C and D and B and A
5 none of the above
Explanation:
B Let naming the diodes \(\mathrm{D}_1, \mathrm{D}_2, \mathrm{D}_3\) and \(\mathrm{D}_4\) According to the question circuit is act as full wave rectifier. In above figure, when we apply \(\mathrm{AC}\) input across \(\mathrm{BD}\). For positive half cycle diode \(D_1 \& D_3\) act as forward bias and \(\mathrm{D}_2 \& \mathrm{D}_4\) act as reverse bias. For negative half cycle \(D_1 \& D_3\) act as a reverse bias and \(D_2 \& D_4\) act as forward bias. Hence, AC input should be connected across B and D and the DC. Output would appear across A and C.
Kerala CEE 2004
Semiconductor Electronics Material Devices and Simple Circuits
150750
In the circuit of figure, treat diode as ideal, current in the \(4 \Omega\) resistor is
1 \(2 \mathrm{~A}\)
2 \(3 \mathrm{~A}\)
3 \(\frac{12}{7} \mathrm{~A}\)
4 \(\frac{30}{13} \mathrm{~A}\)
Explanation:
B From above figure, we can see diode \(\mathrm{D}_1\) is forward bias so, current flow through it and \(\mathrm{D}_2\) is reverse bias and it acts like a open circuit. Then resistance \(2 \Omega\) and \(4 \Omega\) are in series connection. \(\therefore \mathrm{R}_{\mathrm{eq}}=2+4=6 \Omega\) Now, current in \(4 \Omega\) resistance is \(I=\frac{V}{R_{e q}}\) \(I=\frac{12}{6}=2 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150752
A diode is connected to \(220 \mathrm{~V}\) (rms) \(\mathrm{AC}\) in series with a capacitor as shown in figure. The voltage across the capacitor is
1 \(220 \mathrm{~V}\)
2 \(110 \mathrm{~V}\)
3 \(311.1 \mathrm{~V}\)
4 \(\frac{220}{\sqrt{2}} \mathrm{~V}\)
Explanation:
D In this circuit diode D conduct in positive half cycle and in negative half cycle this is open circuit so it behaves as half wave rectifier. The voltage across the capacitor, \(\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}_0}{2}=\frac{220 \sqrt{2}}{2} \Rightarrow \frac{220}{\sqrt{2}}\)
JCECE-2012
Semiconductor Electronics Material Devices and Simple Circuits
150761
In a transistor connected in a common emitter \(\operatorname{mode}, R_C=4 \mathrm{k} \Omega, R_1=1 \mathrm{k} \Omega, I_C=1 \mathrm{~mA}\) and \(I_b=\) \(20 \mu \mathrm{A}\). The voltage gain is
1 100
2 200
3 300
4 400
Explanation:
B Given that, \(\mathrm{R}_{\mathrm{c}}=4 \times 10^3 \Omega \mathrm{R}_1=1 \times 10^3 \Omega\) \(\mathrm{I}_{\mathrm{C}}=1 \times 10^{-3} \mathrm{I}_{\mathrm{B}}=20 \times 10^{-6} \mathrm{~A}\) We know that, voltage gain, \(A_V=\frac{\beta \mathrm{R}_C}{R_1}=\frac{I_C}{I_B} \frac{R_c}{R_1} \quad\left(\because \beta=\frac{I_C}{I_B}\right)\) \(A_v=\frac{1 \times 10^{-3} \times 4 \times 10^3}{20 \times 10^{-6} \times 1 \times 10^3}\) \(A_V=200\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Semiconductor Electronics Material Devices and Simple Circuits
150749
For the given circuit shown below, to act as full wave rectifier, the \(\mathrm{AC}\) input should be connected across and and the DC. output would appear across and
1 B and D and A and C
2 B and A and C and D
3 C and A and B and D
4 C and D and B and A
5 none of the above
Explanation:
B Let naming the diodes \(\mathrm{D}_1, \mathrm{D}_2, \mathrm{D}_3\) and \(\mathrm{D}_4\) According to the question circuit is act as full wave rectifier. In above figure, when we apply \(\mathrm{AC}\) input across \(\mathrm{BD}\). For positive half cycle diode \(D_1 \& D_3\) act as forward bias and \(\mathrm{D}_2 \& \mathrm{D}_4\) act as reverse bias. For negative half cycle \(D_1 \& D_3\) act as a reverse bias and \(D_2 \& D_4\) act as forward bias. Hence, AC input should be connected across B and D and the DC. Output would appear across A and C.
Kerala CEE 2004
Semiconductor Electronics Material Devices and Simple Circuits
150750
In the circuit of figure, treat diode as ideal, current in the \(4 \Omega\) resistor is
1 \(2 \mathrm{~A}\)
2 \(3 \mathrm{~A}\)
3 \(\frac{12}{7} \mathrm{~A}\)
4 \(\frac{30}{13} \mathrm{~A}\)
Explanation:
B From above figure, we can see diode \(\mathrm{D}_1\) is forward bias so, current flow through it and \(\mathrm{D}_2\) is reverse bias and it acts like a open circuit. Then resistance \(2 \Omega\) and \(4 \Omega\) are in series connection. \(\therefore \mathrm{R}_{\mathrm{eq}}=2+4=6 \Omega\) Now, current in \(4 \Omega\) resistance is \(I=\frac{V}{R_{e q}}\) \(I=\frac{12}{6}=2 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150752
A diode is connected to \(220 \mathrm{~V}\) (rms) \(\mathrm{AC}\) in series with a capacitor as shown in figure. The voltage across the capacitor is
1 \(220 \mathrm{~V}\)
2 \(110 \mathrm{~V}\)
3 \(311.1 \mathrm{~V}\)
4 \(\frac{220}{\sqrt{2}} \mathrm{~V}\)
Explanation:
D In this circuit diode D conduct in positive half cycle and in negative half cycle this is open circuit so it behaves as half wave rectifier. The voltage across the capacitor, \(\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}_0}{2}=\frac{220 \sqrt{2}}{2} \Rightarrow \frac{220}{\sqrt{2}}\)
JCECE-2012
Semiconductor Electronics Material Devices and Simple Circuits
150761
In a transistor connected in a common emitter \(\operatorname{mode}, R_C=4 \mathrm{k} \Omega, R_1=1 \mathrm{k} \Omega, I_C=1 \mathrm{~mA}\) and \(I_b=\) \(20 \mu \mathrm{A}\). The voltage gain is
1 100
2 200
3 300
4 400
Explanation:
B Given that, \(\mathrm{R}_{\mathrm{c}}=4 \times 10^3 \Omega \mathrm{R}_1=1 \times 10^3 \Omega\) \(\mathrm{I}_{\mathrm{C}}=1 \times 10^{-3} \mathrm{I}_{\mathrm{B}}=20 \times 10^{-6} \mathrm{~A}\) We know that, voltage gain, \(A_V=\frac{\beta \mathrm{R}_C}{R_1}=\frac{I_C}{I_B} \frac{R_c}{R_1} \quad\left(\because \beta=\frac{I_C}{I_B}\right)\) \(A_v=\frac{1 \times 10^{-3} \times 4 \times 10^3}{20 \times 10^{-6} \times 1 \times 10^3}\) \(A_V=200\)
