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Electric Conductivity MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • Electric Conductivity is considered one the most difficult concept.

  • 26 Questions around this concept.

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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

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

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

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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?

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:

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

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

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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

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

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The transfer ratio of a transistor is 50. The input resistance of the transistor when used in the common emitter configuration is \mathrm{1\ K \Omega}. The peak value for an \mathrm{A.C} input voltage of \mathrm{0.01\ V} of collector current is

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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 μh and the mobility of electron in the same crystal is μe

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

J_{h}=en_h v_{h}=e n_h \mu_{h} E

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

J_{e}=\operatorname{en_ev}_{e}=\operatorname{en_e} \mu_{e} E

hence resultant current density would be

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 \\ and \ \ \ J =\sigma E

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. 

i.e  n_{e}=n_{h}=n_{i} 

 \sigma =n_{i}e (\mu _{e}+ \mu _{h} )

 

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Electric Conductivity

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