Semiconductor Electronics Material Devices and Simple Circuits
150797
For an ideal diode, the current in the following arrangement is
1 \(10 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(10 \mathrm{~A}\)
4 \(1 \mathrm{~mA}\)
Explanation:
B Potential difference across the diode \(=2\) \((-2)=4 \mathrm{~V}\) So the diode is forward bias Now \(\quad \mathrm{V}=\mathrm{IR}\) \(I=\frac{V}{R} \quad(R=400 \Omega)\) \(=\frac{4}{400}\) \(=0.01 \mathrm{~A}\) \(I=10 \mathrm{~mA}\)
MHT-CET 2020
Semiconductor Electronics Material Devices and Simple Circuits
150799
Consider the junction diode as ideal. The value of current flowing through \(A B\) is
1 \(0 \mathrm{~A}\)
2 \(10^{-2} \mathrm{~A}\)
3 \(10^{-1} \mathrm{~A}\)
4 \(10^{-3} \mathrm{~A}\)
Explanation:
B Potential difference across the diode \(=+4-(-6)=10 \mathrm{~V}\) So, the diode is forward bias. Now, \(I=\frac{V}{R}\) \(I=\frac{10}{1 \times 10^3}\) \(I=10^{-2} \mathrm{~A}\)
VITEEE-2018
Semiconductor Electronics Material Devices and Simple Circuits
150796
A p-n junction (D) shown in the figure can act as a rectifier. an alternating current source (V) is connected in the circuit. The current (I) in the resistor( \(R\) ) can be shown by
1
2
3
4
Explanation:
C The given circuit works as half-wave rectifier. In the circuit, we will get current through \(R\) when \(p-n\) junction diode is forward biased and there is no current when p-n junction is reversed biased. Thus the current through resistance will be shown by graph (c).
COMEDK 2018
Semiconductor Electronics Material Devices and Simple Circuits
150800
The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is:
1 10
2 \(10^{-6}\)
3 \(10^6\)
4 100
Explanation:
B The formula for forward bias resistance \(\mathrm{R}_{\mathrm{f}}==\frac{\Delta \mathrm{V}_{\mathrm{f}}}{\Delta \mathrm{I}_{\mathrm{f}}}\) For the given diagram- \(\Delta \mathrm{I}_{\mathrm{f}}=20-10=10 \mathrm{~mA}\) \(\Delta \mathrm{I}_{\mathrm{f}}=10 \times 10^{-3} \mathrm{~A}\) And, \(\Delta \mathrm{V}_{\mathrm{f}} =0.8-0.7\) \(=0.1 \mathrm{~V}\) Hence, \(\mathrm{R}_{\mathrm{f}}=\frac{0.1}{10 \times 10^{-3}} \Omega\) \(\mathrm{R}_{\mathrm{f}}=10 \Omega\) Now, For reverse bias resistance \(R_r\) is given as- \(\mathrm{R}_{\mathrm{r}}=\frac{\Delta \mathrm{V}_{\mathrm{r}}}{\Delta \mathrm{I}_{\mathrm{r}}}\) \(\mathrm{R}_{\mathrm{r}}=\frac{10}{10^{-6}}\) \(\mathrm{R}_{\mathrm{r}}=10^7 \Omega\) \(\therefore\) Ratio resistance \(=\frac{\mathrm{R}_{\mathrm{f}}}{\mathrm{R}_{\mathrm{r}}}=\frac{10}{10^7} \Omega\) \(=10^{-6} \Omega\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Semiconductor Electronics Material Devices and Simple Circuits
150797
For an ideal diode, the current in the following arrangement is
1 \(10 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(10 \mathrm{~A}\)
4 \(1 \mathrm{~mA}\)
Explanation:
B Potential difference across the diode \(=2\) \((-2)=4 \mathrm{~V}\) So the diode is forward bias Now \(\quad \mathrm{V}=\mathrm{IR}\) \(I=\frac{V}{R} \quad(R=400 \Omega)\) \(=\frac{4}{400}\) \(=0.01 \mathrm{~A}\) \(I=10 \mathrm{~mA}\)
MHT-CET 2020
Semiconductor Electronics Material Devices and Simple Circuits
150799
Consider the junction diode as ideal. The value of current flowing through \(A B\) is
1 \(0 \mathrm{~A}\)
2 \(10^{-2} \mathrm{~A}\)
3 \(10^{-1} \mathrm{~A}\)
4 \(10^{-3} \mathrm{~A}\)
Explanation:
B Potential difference across the diode \(=+4-(-6)=10 \mathrm{~V}\) So, the diode is forward bias. Now, \(I=\frac{V}{R}\) \(I=\frac{10}{1 \times 10^3}\) \(I=10^{-2} \mathrm{~A}\)
VITEEE-2018
Semiconductor Electronics Material Devices and Simple Circuits
150796
A p-n junction (D) shown in the figure can act as a rectifier. an alternating current source (V) is connected in the circuit. The current (I) in the resistor( \(R\) ) can be shown by
1
2
3
4
Explanation:
C The given circuit works as half-wave rectifier. In the circuit, we will get current through \(R\) when \(p-n\) junction diode is forward biased and there is no current when p-n junction is reversed biased. Thus the current through resistance will be shown by graph (c).
COMEDK 2018
Semiconductor Electronics Material Devices and Simple Circuits
150800
The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is:
1 10
2 \(10^{-6}\)
3 \(10^6\)
4 100
Explanation:
B The formula for forward bias resistance \(\mathrm{R}_{\mathrm{f}}==\frac{\Delta \mathrm{V}_{\mathrm{f}}}{\Delta \mathrm{I}_{\mathrm{f}}}\) For the given diagram- \(\Delta \mathrm{I}_{\mathrm{f}}=20-10=10 \mathrm{~mA}\) \(\Delta \mathrm{I}_{\mathrm{f}}=10 \times 10^{-3} \mathrm{~A}\) And, \(\Delta \mathrm{V}_{\mathrm{f}} =0.8-0.7\) \(=0.1 \mathrm{~V}\) Hence, \(\mathrm{R}_{\mathrm{f}}=\frac{0.1}{10 \times 10^{-3}} \Omega\) \(\mathrm{R}_{\mathrm{f}}=10 \Omega\) Now, For reverse bias resistance \(R_r\) is given as- \(\mathrm{R}_{\mathrm{r}}=\frac{\Delta \mathrm{V}_{\mathrm{r}}}{\Delta \mathrm{I}_{\mathrm{r}}}\) \(\mathrm{R}_{\mathrm{r}}=\frac{10}{10^{-6}}\) \(\mathrm{R}_{\mathrm{r}}=10^7 \Omega\) \(\therefore\) Ratio resistance \(=\frac{\mathrm{R}_{\mathrm{f}}}{\mathrm{R}_{\mathrm{r}}}=\frac{10}{10^7} \Omega\) \(=10^{-6} \Omega\)
Semiconductor Electronics Material Devices and Simple Circuits
150797
For an ideal diode, the current in the following arrangement is
1 \(10 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(10 \mathrm{~A}\)
4 \(1 \mathrm{~mA}\)
Explanation:
B Potential difference across the diode \(=2\) \((-2)=4 \mathrm{~V}\) So the diode is forward bias Now \(\quad \mathrm{V}=\mathrm{IR}\) \(I=\frac{V}{R} \quad(R=400 \Omega)\) \(=\frac{4}{400}\) \(=0.01 \mathrm{~A}\) \(I=10 \mathrm{~mA}\)
MHT-CET 2020
Semiconductor Electronics Material Devices and Simple Circuits
150799
Consider the junction diode as ideal. The value of current flowing through \(A B\) is
1 \(0 \mathrm{~A}\)
2 \(10^{-2} \mathrm{~A}\)
3 \(10^{-1} \mathrm{~A}\)
4 \(10^{-3} \mathrm{~A}\)
Explanation:
B Potential difference across the diode \(=+4-(-6)=10 \mathrm{~V}\) So, the diode is forward bias. Now, \(I=\frac{V}{R}\) \(I=\frac{10}{1 \times 10^3}\) \(I=10^{-2} \mathrm{~A}\)
VITEEE-2018
Semiconductor Electronics Material Devices and Simple Circuits
150796
A p-n junction (D) shown in the figure can act as a rectifier. an alternating current source (V) is connected in the circuit. The current (I) in the resistor( \(R\) ) can be shown by
1
2
3
4
Explanation:
C The given circuit works as half-wave rectifier. In the circuit, we will get current through \(R\) when \(p-n\) junction diode is forward biased and there is no current when p-n junction is reversed biased. Thus the current through resistance will be shown by graph (c).
COMEDK 2018
Semiconductor Electronics Material Devices and Simple Circuits
150800
The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is:
1 10
2 \(10^{-6}\)
3 \(10^6\)
4 100
Explanation:
B The formula for forward bias resistance \(\mathrm{R}_{\mathrm{f}}==\frac{\Delta \mathrm{V}_{\mathrm{f}}}{\Delta \mathrm{I}_{\mathrm{f}}}\) For the given diagram- \(\Delta \mathrm{I}_{\mathrm{f}}=20-10=10 \mathrm{~mA}\) \(\Delta \mathrm{I}_{\mathrm{f}}=10 \times 10^{-3} \mathrm{~A}\) And, \(\Delta \mathrm{V}_{\mathrm{f}} =0.8-0.7\) \(=0.1 \mathrm{~V}\) Hence, \(\mathrm{R}_{\mathrm{f}}=\frac{0.1}{10 \times 10^{-3}} \Omega\) \(\mathrm{R}_{\mathrm{f}}=10 \Omega\) Now, For reverse bias resistance \(R_r\) is given as- \(\mathrm{R}_{\mathrm{r}}=\frac{\Delta \mathrm{V}_{\mathrm{r}}}{\Delta \mathrm{I}_{\mathrm{r}}}\) \(\mathrm{R}_{\mathrm{r}}=\frac{10}{10^{-6}}\) \(\mathrm{R}_{\mathrm{r}}=10^7 \Omega\) \(\therefore\) Ratio resistance \(=\frac{\mathrm{R}_{\mathrm{f}}}{\mathrm{R}_{\mathrm{r}}}=\frac{10}{10^7} \Omega\) \(=10^{-6} \Omega\)
Semiconductor Electronics Material Devices and Simple Circuits
150797
For an ideal diode, the current in the following arrangement is
1 \(10 \mathrm{~mA}\)
2 \(20 \mathrm{~mA}\)
3 \(10 \mathrm{~A}\)
4 \(1 \mathrm{~mA}\)
Explanation:
B Potential difference across the diode \(=2\) \((-2)=4 \mathrm{~V}\) So the diode is forward bias Now \(\quad \mathrm{V}=\mathrm{IR}\) \(I=\frac{V}{R} \quad(R=400 \Omega)\) \(=\frac{4}{400}\) \(=0.01 \mathrm{~A}\) \(I=10 \mathrm{~mA}\)
MHT-CET 2020
Semiconductor Electronics Material Devices and Simple Circuits
150799
Consider the junction diode as ideal. The value of current flowing through \(A B\) is
1 \(0 \mathrm{~A}\)
2 \(10^{-2} \mathrm{~A}\)
3 \(10^{-1} \mathrm{~A}\)
4 \(10^{-3} \mathrm{~A}\)
Explanation:
B Potential difference across the diode \(=+4-(-6)=10 \mathrm{~V}\) So, the diode is forward bias. Now, \(I=\frac{V}{R}\) \(I=\frac{10}{1 \times 10^3}\) \(I=10^{-2} \mathrm{~A}\)
VITEEE-2018
Semiconductor Electronics Material Devices and Simple Circuits
150796
A p-n junction (D) shown in the figure can act as a rectifier. an alternating current source (V) is connected in the circuit. The current (I) in the resistor( \(R\) ) can be shown by
1
2
3
4
Explanation:
C The given circuit works as half-wave rectifier. In the circuit, we will get current through \(R\) when \(p-n\) junction diode is forward biased and there is no current when p-n junction is reversed biased. Thus the current through resistance will be shown by graph (c).
COMEDK 2018
Semiconductor Electronics Material Devices and Simple Circuits
150800
The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is:
1 10
2 \(10^{-6}\)
3 \(10^6\)
4 100
Explanation:
B The formula for forward bias resistance \(\mathrm{R}_{\mathrm{f}}==\frac{\Delta \mathrm{V}_{\mathrm{f}}}{\Delta \mathrm{I}_{\mathrm{f}}}\) For the given diagram- \(\Delta \mathrm{I}_{\mathrm{f}}=20-10=10 \mathrm{~mA}\) \(\Delta \mathrm{I}_{\mathrm{f}}=10 \times 10^{-3} \mathrm{~A}\) And, \(\Delta \mathrm{V}_{\mathrm{f}} =0.8-0.7\) \(=0.1 \mathrm{~V}\) Hence, \(\mathrm{R}_{\mathrm{f}}=\frac{0.1}{10 \times 10^{-3}} \Omega\) \(\mathrm{R}_{\mathrm{f}}=10 \Omega\) Now, For reverse bias resistance \(R_r\) is given as- \(\mathrm{R}_{\mathrm{r}}=\frac{\Delta \mathrm{V}_{\mathrm{r}}}{\Delta \mathrm{I}_{\mathrm{r}}}\) \(\mathrm{R}_{\mathrm{r}}=\frac{10}{10^{-6}}\) \(\mathrm{R}_{\mathrm{r}}=10^7 \Omega\) \(\therefore\) Ratio resistance \(=\frac{\mathrm{R}_{\mathrm{f}}}{\mathrm{R}_{\mathrm{r}}}=\frac{10}{10^7} \Omega\) \(=10^{-6} \Omega\)
