Semiconductor Electronics Material Devices and Simple Circuits
150802
Current in the circuit will be
1 \(\frac{5}{40} \mathrm{~A}\)
2 \(\frac{5}{50} \mathrm{~A}\)
3 \(\frac{5}{10} \mathrm{~A}\)
4 \(\frac{5}{20} \mathrm{~A}\)
Explanation:
B From the above figure \(\mathrm{D}_1 \rightarrow\) reverse biased \(\mathrm{D}_2 \rightarrow\) forward biased So, no current will be flow through \(\mathrm{D}_1\) diode Hence, Equivalent circuit is- So, \(I=\frac{5 V}{(30+20) \Omega}=\frac{5}{50} A\)
VITEEE-2014
Semiconductor Electronics Material Devices and Simple Circuits
150806
A \(220 \mathrm{~V}\) AC supply is connected between points \(A\) and \(B\) as shown in figure, what will be the potential difference \(V\) across the capacitor?
1 \(220 \mathrm{~V}\)
2 \(110 \mathrm{~V}\)
3 zero
4 \(220 \sqrt{2} \mathrm{~V}\)
Explanation:
D During (+ve) half cycle, p-n junction diode conducts. So, capacitor is charged during \((+)\) ve half cycle. Now, \(\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}}{\sqrt{2}}\) \(\mathrm{~V}=\mathrm{V}_{\mathrm{rms}} \sqrt{2}\) \(\mathrm{~V}=220 \sqrt{2} \mathrm{~V}\)
Karnataka CET-2020
Semiconductor Electronics Material Devices and Simple Circuits
150807
In the figure shown, if the diode forward voltage drop is \(0.2 \mathrm{~V}\), the voltage difference between \(A\) and \(B\) is :
Semiconductor Electronics Material Devices and Simple Circuits
150808
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
1 \(2.31 \mathrm{~A}\)
2 \(1.71 \mathrm{~A}\)
3 \(1.33 \mathrm{~A}\)
4 \(2.0 \mathrm{~A}\)
Explanation:
B Given that, Diode \(\mathrm{D}_2\) is in reverse bias \& \(\mathrm{D}_1\) is in forward biased. \(\therefore \quad\) New circuit, Apply KVL- \(12-4 \mathrm{I}-3 \mathrm{I}=0\) \(7 \mathrm{I}=12\) \(\mathrm{I}=\frac{12}{7}=1.71\) \(\mathrm{I}=1.71 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150802
Current in the circuit will be
1 \(\frac{5}{40} \mathrm{~A}\)
2 \(\frac{5}{50} \mathrm{~A}\)
3 \(\frac{5}{10} \mathrm{~A}\)
4 \(\frac{5}{20} \mathrm{~A}\)
Explanation:
B From the above figure \(\mathrm{D}_1 \rightarrow\) reverse biased \(\mathrm{D}_2 \rightarrow\) forward biased So, no current will be flow through \(\mathrm{D}_1\) diode Hence, Equivalent circuit is- So, \(I=\frac{5 V}{(30+20) \Omega}=\frac{5}{50} A\)
VITEEE-2014
Semiconductor Electronics Material Devices and Simple Circuits
150806
A \(220 \mathrm{~V}\) AC supply is connected between points \(A\) and \(B\) as shown in figure, what will be the potential difference \(V\) across the capacitor?
1 \(220 \mathrm{~V}\)
2 \(110 \mathrm{~V}\)
3 zero
4 \(220 \sqrt{2} \mathrm{~V}\)
Explanation:
D During (+ve) half cycle, p-n junction diode conducts. So, capacitor is charged during \((+)\) ve half cycle. Now, \(\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}}{\sqrt{2}}\) \(\mathrm{~V}=\mathrm{V}_{\mathrm{rms}} \sqrt{2}\) \(\mathrm{~V}=220 \sqrt{2} \mathrm{~V}\)
Karnataka CET-2020
Semiconductor Electronics Material Devices and Simple Circuits
150807
In the figure shown, if the diode forward voltage drop is \(0.2 \mathrm{~V}\), the voltage difference between \(A\) and \(B\) is :
Semiconductor Electronics Material Devices and Simple Circuits
150808
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
1 \(2.31 \mathrm{~A}\)
2 \(1.71 \mathrm{~A}\)
3 \(1.33 \mathrm{~A}\)
4 \(2.0 \mathrm{~A}\)
Explanation:
B Given that, Diode \(\mathrm{D}_2\) is in reverse bias \& \(\mathrm{D}_1\) is in forward biased. \(\therefore \quad\) New circuit, Apply KVL- \(12-4 \mathrm{I}-3 \mathrm{I}=0\) \(7 \mathrm{I}=12\) \(\mathrm{I}=\frac{12}{7}=1.71\) \(\mathrm{I}=1.71 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150802
Current in the circuit will be
1 \(\frac{5}{40} \mathrm{~A}\)
2 \(\frac{5}{50} \mathrm{~A}\)
3 \(\frac{5}{10} \mathrm{~A}\)
4 \(\frac{5}{20} \mathrm{~A}\)
Explanation:
B From the above figure \(\mathrm{D}_1 \rightarrow\) reverse biased \(\mathrm{D}_2 \rightarrow\) forward biased So, no current will be flow through \(\mathrm{D}_1\) diode Hence, Equivalent circuit is- So, \(I=\frac{5 V}{(30+20) \Omega}=\frac{5}{50} A\)
VITEEE-2014
Semiconductor Electronics Material Devices and Simple Circuits
150806
A \(220 \mathrm{~V}\) AC supply is connected between points \(A\) and \(B\) as shown in figure, what will be the potential difference \(V\) across the capacitor?
1 \(220 \mathrm{~V}\)
2 \(110 \mathrm{~V}\)
3 zero
4 \(220 \sqrt{2} \mathrm{~V}\)
Explanation:
D During (+ve) half cycle, p-n junction diode conducts. So, capacitor is charged during \((+)\) ve half cycle. Now, \(\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}}{\sqrt{2}}\) \(\mathrm{~V}=\mathrm{V}_{\mathrm{rms}} \sqrt{2}\) \(\mathrm{~V}=220 \sqrt{2} \mathrm{~V}\)
Karnataka CET-2020
Semiconductor Electronics Material Devices and Simple Circuits
150807
In the figure shown, if the diode forward voltage drop is \(0.2 \mathrm{~V}\), the voltage difference between \(A\) and \(B\) is :
Semiconductor Electronics Material Devices and Simple Circuits
150808
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
1 \(2.31 \mathrm{~A}\)
2 \(1.71 \mathrm{~A}\)
3 \(1.33 \mathrm{~A}\)
4 \(2.0 \mathrm{~A}\)
Explanation:
B Given that, Diode \(\mathrm{D}_2\) is in reverse bias \& \(\mathrm{D}_1\) is in forward biased. \(\therefore \quad\) New circuit, Apply KVL- \(12-4 \mathrm{I}-3 \mathrm{I}=0\) \(7 \mathrm{I}=12\) \(\mathrm{I}=\frac{12}{7}=1.71\) \(\mathrm{I}=1.71 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150802
Current in the circuit will be
1 \(\frac{5}{40} \mathrm{~A}\)
2 \(\frac{5}{50} \mathrm{~A}\)
3 \(\frac{5}{10} \mathrm{~A}\)
4 \(\frac{5}{20} \mathrm{~A}\)
Explanation:
B From the above figure \(\mathrm{D}_1 \rightarrow\) reverse biased \(\mathrm{D}_2 \rightarrow\) forward biased So, no current will be flow through \(\mathrm{D}_1\) diode Hence, Equivalent circuit is- So, \(I=\frac{5 V}{(30+20) \Omega}=\frac{5}{50} A\)
VITEEE-2014
Semiconductor Electronics Material Devices and Simple Circuits
150806
A \(220 \mathrm{~V}\) AC supply is connected between points \(A\) and \(B\) as shown in figure, what will be the potential difference \(V\) across the capacitor?
1 \(220 \mathrm{~V}\)
2 \(110 \mathrm{~V}\)
3 zero
4 \(220 \sqrt{2} \mathrm{~V}\)
Explanation:
D During (+ve) half cycle, p-n junction diode conducts. So, capacitor is charged during \((+)\) ve half cycle. Now, \(\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}}{\sqrt{2}}\) \(\mathrm{~V}=\mathrm{V}_{\mathrm{rms}} \sqrt{2}\) \(\mathrm{~V}=220 \sqrt{2} \mathrm{~V}\)
Karnataka CET-2020
Semiconductor Electronics Material Devices and Simple Circuits
150807
In the figure shown, if the diode forward voltage drop is \(0.2 \mathrm{~V}\), the voltage difference between \(A\) and \(B\) is :
Semiconductor Electronics Material Devices and Simple Circuits
150808
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
1 \(2.31 \mathrm{~A}\)
2 \(1.71 \mathrm{~A}\)
3 \(1.33 \mathrm{~A}\)
4 \(2.0 \mathrm{~A}\)
Explanation:
B Given that, Diode \(\mathrm{D}_2\) is in reverse bias \& \(\mathrm{D}_1\) is in forward biased. \(\therefore \quad\) New circuit, Apply KVL- \(12-4 \mathrm{I}-3 \mathrm{I}=0\) \(7 \mathrm{I}=12\) \(\mathrm{I}=\frac{12}{7}=1.71\) \(\mathrm{I}=1.71 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150802
Current in the circuit will be
1 \(\frac{5}{40} \mathrm{~A}\)
2 \(\frac{5}{50} \mathrm{~A}\)
3 \(\frac{5}{10} \mathrm{~A}\)
4 \(\frac{5}{20} \mathrm{~A}\)
Explanation:
B From the above figure \(\mathrm{D}_1 \rightarrow\) reverse biased \(\mathrm{D}_2 \rightarrow\) forward biased So, no current will be flow through \(\mathrm{D}_1\) diode Hence, Equivalent circuit is- So, \(I=\frac{5 V}{(30+20) \Omega}=\frac{5}{50} A\)
VITEEE-2014
Semiconductor Electronics Material Devices and Simple Circuits
150806
A \(220 \mathrm{~V}\) AC supply is connected between points \(A\) and \(B\) as shown in figure, what will be the potential difference \(V\) across the capacitor?
1 \(220 \mathrm{~V}\)
2 \(110 \mathrm{~V}\)
3 zero
4 \(220 \sqrt{2} \mathrm{~V}\)
Explanation:
D During (+ve) half cycle, p-n junction diode conducts. So, capacitor is charged during \((+)\) ve half cycle. Now, \(\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}}{\sqrt{2}}\) \(\mathrm{~V}=\mathrm{V}_{\mathrm{rms}} \sqrt{2}\) \(\mathrm{~V}=220 \sqrt{2} \mathrm{~V}\)
Karnataka CET-2020
Semiconductor Electronics Material Devices and Simple Circuits
150807
In the figure shown, if the diode forward voltage drop is \(0.2 \mathrm{~V}\), the voltage difference between \(A\) and \(B\) is :
Semiconductor Electronics Material Devices and Simple Circuits
150808
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
1 \(2.31 \mathrm{~A}\)
2 \(1.71 \mathrm{~A}\)
3 \(1.33 \mathrm{~A}\)
4 \(2.0 \mathrm{~A}\)
Explanation:
B Given that, Diode \(\mathrm{D}_2\) is in reverse bias \& \(\mathrm{D}_1\) is in forward biased. \(\therefore \quad\) New circuit, Apply KVL- \(12-4 \mathrm{I}-3 \mathrm{I}=0\) \(7 \mathrm{I}=12\) \(\mathrm{I}=\frac{12}{7}=1.71\) \(\mathrm{I}=1.71 \mathrm{~A}\)
