Physics 5 Marks Questions

1. The change in output voltage is given by

$\text{V}_0=\text{I}_\text{c}\times\text{R}_\text{L}$

$\because\text{I}_\text{c}=\beta\times\text{I}_\text{b}$

$\therefore\text{V}_0=\beta\times\text{I}_\text{b}\times\text{R}_\text{L}$

$\Rightarrow\text{V}_0=50\times50\mu\text{A}\times2\text{k}\Omega$

$\Rightarrow\text{V}_0=5\text{V}$

2. The change in input voltage is given by

$\delta\text{V}_\text{i}=\delta\text{l}_\text{b}\times\text{R}_\text{i}$

$\Rightarrow\delta\text{V}_\text{i}=50\times10^{-6}\times5\times10^2$

$\Rightarrow\delta\text{V}_\text{i}=25\times10^{-3}$

$\Rightarrow\delta\text{V}_\text{i}=25\text{mV}$

3. Power gain is given by

$\beta^2\times\frac{\text{R}_\text{L}}{\text{R}_\text{i}}$

$\Rightarrow2500\times\frac{2}{0.5}$

$\Rightarrow2500\times\frac{20}{5}=10^4.$

1. $\text{A}=1,\text{B}=0,\text{C}=1$

$\text{X}=1.\Big(\overline{0.1}\Big)+0\Big(\overline{1.1}\Big)+1\Big(\overline1.0\Big)$

$=1.\bar{0}+0.\bar{1}+1.\bar{0}$

$=1.1+0.0+1.1$

$=1+0+1$

$=1+1$

$=1$

2. $\text{A}=\text{B}=\text{C}=1$

$\text{X}=1\Big(\overline{1.1}\Big)+1\Big(\overline{1.1}\Big)+1\Big(\overline{1.1}\Big)$

$=1.\bar{1}+1.\bar{1}+1.\bar{1}$

$=1.0+1.0+1.0$

$=0+0+0$

$=0$

3. $\text{A}=\text{B}=\text{C}=0$

$\text{X}=0\Big(\overline{0.0}\Big)+0\Big(\overline{0.0}\Big)+0\Big(\overline{1.1}\Big)$

$=0.\bar{0}+0.\bar{0}+0.\bar{0}$

$=0.1+0.1+0.1$

$=0+0+0$

$=0.$

1. $\text{i}_0=20\times10^{-6}\text{A},\ \text{T}=300\text{K, V}=300\text{mV}$

$\text{i}=\text{i}_0\text{e}^{\frac{\text{eV}}{\text{KT}}-1}$ $=20\times10^{-6}\Big(\text{e}^{\frac{100}{8.62}}-1\Big)=2.18\text{A}=2\text{A}.$

2. $4=20\times10^{-6}\Big(\text{e}^{\frac{\text{V}}{8.62\times3\times10^{-2}}}-1\Big)$

$\Rightarrow\text{e}^{\frac{\text{V}\times10^3}{8.62\times3}}-1=\frac{4\times10^6}{20}$

$\Rightarrow\text{e}^{\frac{\text{V}\times10^3}{8.62\times3}}=200001\Rightarrow\frac{\text{V}\times10^{3}}{8.62\times3}=12.2060$

$\Rightarrow\text{V}=315\text{mV}=318\text{mV}.$

1. $\text{i}_1=25\mu\text{A},\text{V}=200\text{mV},\text{i}_2=75\mu\text{A}$

When in unbiased condition drift current = diffusion current

$\therefore$ Diffusion current $=25\mu\text{A}.$

2. On reverse biasing the diffusion current becomes ‘O’.
3. On forward biasing the actual current be x.

x - Drift current = Forward biasing current

$\Rightarrow\text{x}-25\mu\text{A}=75\mu\text{A}$

$\Rightarrow\text{x}=(75+25)\mu\text{A}=100\mu\text{A}.$

1. When $\text{R}=12\Omega$

The wire EF becomes ineffective due to the net (-)ve voltage.

Hence, current through $\text{R}=\frac{10}{24}=0.4166=042\text{A}.$

2. Similarly for $\text{R}=48\Omega$

$\text{i}=\frac{10}{(48+12)}=\frac{10}{60}=0.16\text{A}.$

1. The current‒voltage relationship of a diode is given by $\text{i}=\text{i}_0\Big(\text{e}^{\frac{\text{eV}}{\text{kT}-1}}\Big)$

For a large value of voltage, 1 can be neglected.

$\text{i}\approx\text{i}_0\text{e}^{\frac{\text{eV}}{\text{KT}}}$

Again, we need to find the voltage at which

$\text{e}^{\frac{\text{eV}}{\text{KT}}}=100$

$\Rightarrow\frac{\text{eV}}{\text{kT}}=\text{In }100$

$\Rightarrow\text{V}=\frac{\text{In }100\times\text{kT}}{\text{e}}$

$\Rightarrow\text{V}=\frac{2.303\times\log\ 100\times8.62\times10^{-5}\times300}{\text{e}}$

$\Rightarrow\text{V}=0.12\text{V}$

2. Given:

$\text{i}=\text{i}_0\Big(\text{e}^{\frac{\text{eV}}{\text{kT}-1}}\Big)\ ...(1)$

We know that the dynamic resistance of a diode is the rate of change of voltage w.r.t. current.

i.e. $\text{R}=\frac{\text{dV}}{\text{di}}$

As the exponential factor dominates the factor of 1, we can neglect this factor.

Now, on differentiating eq. (1) w.r.t. V, we get,

$\frac{\text{di}}{\text{dV}}=\text{i}_0\frac{\text{e}}{\text{kT}}\text{e}^{\frac{\text{eV}}{\text{kT}}}$

$\Rightarrow\frac{1}{\text{R}}=\frac{\text{ei}_0}{\text{kT}}\text{e}^{\frac{\text{eV}}{\text{kT}}}$

$\Rightarrow\text{R}=\frac{\text{kT}}{\text{ei}_0}\text{e}^{\frac{-\text{eV}}{\text{kT}}}\ ...(2)$

3.  Given,

$\text{R}=2\Omega$

On substituting this value in eq. (2), we get

$2=\frac{8.62\times10^{-5}\times300}{\text{e}\times10\times10^{-6}}\text{e}^{\frac{\text{-eV}}{8.62\times10^{-5}\times300}}$

$\Rightarrow\text{V}=0.25\text{V}.$

1. Since both the diodes are forward biased net resistance = 0

$\text{i}=\frac{2\text{V}}{2\Omega}=1\text{A}$

2. One of the diodes is forward biased and other is reverse biase.

Thus the resistance of one becomes $\infty.$

$\text{i}=\frac{2}{2+\infty}=0\text{A}$

Both are forward biased.

Thus the resistance is 0.

$\text{i}=\frac{2}{2}=1\text{A}.$

One is forward biased and other is reverse biased.

Thus the current passes through the forward biased diode.

$\therefore\text{i}=\frac{2}{2}=1\text{A}.$