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Electric Conductivity - Practice Questions & MCQ

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

Quick Facts

  • Electric Conductivity is considered one the most difficult concept.

  • 33 Questions around this concept.

Solve by difficulty

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

Following is the relation between current and charge \mathrm{I=A T^2 e^{q / V_L}} , then value of \mathrm{ V_L} 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

If \mathrm{RP = 7\ k\Omega , gm = 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

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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

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

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In the grid circuit of a triode a signal E=2\sqrt{2} cos ωt is applied. If \mu =14 and \mathrm{r_{p}=10\ k\Omega } then root mean square current flowing through \mathrm{R_{L}=12\ k\Omega } will be

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 output signal voltage will be

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 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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