In presence of interspace charge, at a plate voltage of $\mathrm{200\ V}$ , the current is $\mathrm{80\ mA}$ . Then the current in mA at $\mathrm{400\ V}$ will be

$160 \sqrt{2}$

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$2 \sqrt{2}$

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$80 / \sqrt{2}$

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None of these

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Following is the relation between current and charge $\mathrm{I}=\mathrm{AT}^2 \mathrm{e}^{\mathrm{q} / \mathrm{V}_{\mathrm{L}}}$, then value of $\mathrm{V}_{\mathrm{L}}$ will be

$\frac{\mathrm{V}}{\mathrm{kt}}$

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$\frac{\mathrm{KV}}{\mathrm{t}}$

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$\frac{\mathrm{kT}}{\mathrm{V}}$

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$\frac{\mathrm{Vt}}{\mathrm{k}}$

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In a triode valve, the amplification factor is $20$ and mutual conductance is $\mathrm{10^{-3}mho}$ . The plate resistance is

$2 \times 10^3 \Omega$

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$4 \times 10^3 \Omega$

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$2 \times 10^4 \Omega$

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$2 \times 10^5 \Omega$

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If in a triode valve amplification factor is $20$ and plate resistance is $\mathrm{10\ k\Omega }$ , then its mutual conductance is

$\mathrm{2\ milli\ mho}$

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$\mathrm{20\ milli\ mho}$

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$\mathrm{(1/2)\ milli\ mho}$

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$\mathrm{200\ milli\ mho}$

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If $\mathrm{R_P = 7\ k\Omega , g_m = 2.5\ milli\ mho}$ , then on increasing plate voltage by $\mathrm{50\ V}$ , how much the grid voltage is changed so that plate current remains the same?

$\mathrm{-2.86\ V}$

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$\mathrm{-4\ V}$

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$\mathrm{+4\ V}$

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$\mathrm{+2\ V}$

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The value of current in a triode valve is given by $\mathrm{I_p=0.004\left(V_p+10 V_{\mathrm{g}}\right)^{3 / 2} \mathrm{~mA}}$ . When plate potential and grid potential are $\mathrm{120\ V}$ and $\mathrm{-2\ V}$ respectively, then the value of mutual conductance will be:

$\mathrm{600\ mho}$

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$\mathrm{60\ mho}$

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$\mathrm{6\ mho}$

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$\mathrm{6 \times 10^{-4} mho}$

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The amplification factor of a triode valve is $15$ . If the grid voltage is changed by $0.3$ volt the change in plate voltage in order to keep the plate current constant (in volt) is

$0.02$

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$0.002$

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$4.5$

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$5.0$

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For a thermionic emitter (metallic) if J represents the current density and T is its absolute temperature then the correct curve between $\log _{\mathrm{e}} \frac{\mathrm{J}}{\mathrm{T}^2}$ and $\frac{1}{\mathrm{~T}}$ is

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Correct Answer

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In a triode amplifier, $\mathrm{\mu = 25, r_{p} = 40\ kilo\ ohm}$ and load resistance $\mathrm{RL = 10\ kilo\ ohm}$ . If the input signal voltage is $\mathrm{0.5\ volt}$ , then the output signal voltage will be

$\mathrm{1.25\ volt}$

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$\mathrm{5\ volt}$

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$\mathrm{2.5\ volt}$

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$\mathrm{10\ volt}$

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Mutual conductance of the triode is $\mathrm{2\ m \Omega ^{-1}}$ and the amplification factor is $\mathrm{50}$ . Its anode is connected with a source of $\mathrm{250\ V}$ and a resistance of $\mathrm{25\ k \Omega}$ . The voltage gain of this amplifier is

$12.5$

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$10$

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$25$

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$50$

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Concepts Covered - 1

Electric Conductivity

Electrical Conductivity (σ)-

The semiconductor conducts electricity with the help of these two types of electricity or charge carriers (i.e electrons and holes).

These holes and electrons move in the opposite direction. The electrons always tend to move in opposite direction to the applied electric field.

Let the mobility of the hole in the crystal is μ_h and the mobility of electron in the same crystal is μ_e

The current density due to drift of holes is given by,

$$
J_h=e n_h v_h=e n_h \mu_h E
$$

And The current density due to the drift of electrons is given by,

$$
J_e=\mathrm{en}_{\mathrm{e}} \mathrm{v}_e=\mathrm{en}_{\mathrm{e}} \mu_e E
$$

hence resultant current density would be

$$
\begin{aligned}
& \quad J=J_h+J_e=e n_h v_h+e n_e v_e=e n_h \mu_h E+e n_e \mu_e E=\left(n_h \mu_h+n_e \mu_e\right) e E \\
& \text { and } J=\sigma E
\end{aligned}
$$

So, the general equation for conductivity is given as

$$
\sigma=e\left(n_e \mu_e+n_h \mu_h\right)
$$

where
$n_e=$ electron density
$n_h=$ hole density
$\mu_e=$ mobility of electron
$\mu_h=$ mobility of holes

For intrinsic semiconductors (no impurities)-

As the number of electrons will be equal to the number of holes.

$\begin{aligned} & \text { i.e } n_e=n_h=n_i \\ & \sigma=n_i e\left(\mu_e+\mu_h\right)\end{aligned}$