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The wavelength and intensity of light emitted by a LED depend upon
Option: 1  forward bias and energy gap of the semiconductor
Option: 2 energy gap of the semiconductor and reverse bias  
Option: 3 energy gap only 
Option: 4 forward bias only

The wavelength and intensity of light emitted by a LED depend upon  forward bias and energy gap of the semiconductor

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Safeer PP

The wavelength and intensity of light emitted by a LED depend upon 
Option: 1  forward bias and energy gap of the semiconductor 
Option: 2 energy gap of the semiconductor and reverse bias 
Option: 3 energy gap only 
Option: 4 forward bias only

The wavelength and intensity of light emitted by a LED depend upon forward bias and energy gap of the semiconductor 

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Posted by

Safeer PP

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A zener diode has 
Option: 1 heavily doped p-side and lightly doped n-side. 
Option: 2 heavily doped n-side and lightly doped p-side. 
Option: 3 heavily doped n-side as well as p-side. 
Option: 4 lightly doped n-side as well as p-side

A Zener diode has heavily doped n-side as well as p-side. 

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Safeer PP

(a) Draw the graph of radius of orbit (rn) in hydrogen atom as a function of orbit number (n).
(b) In a hydrogen atom, find the ratio of the time taken by the electron to complete one revolution in the first excited and in the second excited states.
Option: 1
Option: 2
Option: 3
Option: 4

a) 

b)

\\r_n\propto n^2\\v_n\propto1/n\\T=\frac{2\pi r_v}{v_n}\\T\propto n^3

\frac{T_2}{T_3}=\frac{8}{27}

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Safeer PP

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In an n-type semiconductor, the donor energy level lies
Option: 1 at the centre of the energy gap.
Option: 2 just below the conduction band.
Option: 3 just above the valance band.
Option: 4 in the conduction band.

In an n-type semiconductor, the donor energy level lies just below the conduction band

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At equilibrium, in a p-n junction diode the net current is
Option: 1 due to diffusion of majority charge carriers.
Option: 2 due to drift of minority charge carriers.
Option: 3 zero as diffusion and drift currents are equal and opposite.
Option: 4 zero as no charge carriers cross the junction.

At equilibrium, in a p-n junction diode, the net current is zero as diffusion and drift current is equal and opposite.

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Safeer PP

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In the circuit shown in Fig.14.20, find the value of RC.

 

Let us consider the ciruit diagram to solve this problem

I_{E}=I_{C}+I_{B}\: and\: I_{C}=\beta I_{B}.........(i)\\ I_{C}R_{C}+V_{CE}+I_{E}R_{E}=V_{CC}..........(ii)\\ RI_{B}+V_{BE}+I_{E}R_{E}=V_{CC}.................(iii)\\ \because I_{E}\approx I_{C}=\beta I_{B}\\ from(iii)\\ \left ( R+\beta R_{E} \right )I_{B}=v_{CC}-V_{BE}\\ \Rightarrow I_{B}=\frac{V_{CC}-V_{BE}}{R+\beta .R_{E}}\\ =\frac{12-0.5}{80+1.2\times 100}=\frac{11.5}{200}mA\\ from(ii)\\ \left ( R_{C}+R_{E} \right )=\frac{V_{CE}-V_{BE}}{I_{C}}=\frac{V_{CC}-V_{CE}}{\beta I_{B}}\: \: \: \: \: \: \: \: \: \left ( \because I_{C}=\beta I_{B} \right )

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For the transistor circuit shown in Fig.14.19, evaluate VE, RB, RE given IC = 1 mA, VCE = 3V, VBE = 0.5 V and VCC = 12 V, \beta = 100.

As we know the base current is very small So,

I_{C}=I_{E}\\ R_{C}=7.8 k\Omega\\ from \, th\, fig. I_{C}\left ( R_{c}+R_{E} \right )+V_{CE}=12\\ \left ( R_{E}+R_{C} \right )\times 1 \times 10^{-3}+3=12\\ \left ( R_{E}+R_{C} \right )=9 \times 10^{3}=9k\Omega\\ R_{E}=9-7.8=1.2k\Omega\\ V_{E}=I_{E}\times R_{E}\\ =1 \times 10^{-3} \times 1.2 \times 10^{3}=1.2V\\ Voltage V_{B}=V_{E}+V_{BE}=1.2+0.5=1.7V\\ Current \: I=\frac{V_{B}}{20 \times 10^{3}}=\frac{1.7}{20 \times 10^{3}}=0.085mA\\ Resistance \: R_{B}=\frac{12-1.7}{\frac{I_{C}}{\beta }+0.085}\times 10^{3}=\frac{10.3}{0.01+0.085} \: \: \: \: \: \: \left [ Given \beta =100 \right ]\\ =108 k\Omega\\

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Consider a box with three terminals on top of it as shown in Fig.14.18 (a):


Three components namely, two germanium diodes and one resistor are connected across these three terminals in some arrangement.
A student performs an experiment in which any two of these three terminals are connected in the circuit shown in Fig. 14.18 (b).

The student obtains graphs of current-voltage characteristics for unknown combination of components between the two terminals connected in the circuit.
The graphs are
(i) when A is positive and B is negative

(ii) when A is negative and B is positive

(iii) When B is negative and C is positive

(iv) When B is positive and C is negative

(v) When A is positive and C is negative

(vi) When A is negative and C is positive

From these graphs of current – voltage characteristic shown in Fig. 14.18 (c) to (h), determine the arrangement of components between A, B and C.

a) n-side of the PN junction is connected to the A terminals while B is connected to the top-side of PN junction.

b) the knee voltage is 0.7V.

c) C is connected to the PN junction and n-side of the junction is connected to the B terminal, and knee voltage is 0.7V.

d) This above condition explains the connection between pn-junction with I and II with the resistance.

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 An X-OR gate has following truth table:

A B Y
0 0 0
0 1 1
1 0 1
1 1 0

It is represented by following logic relation
Y=\bar{A}.B+A.\bar{B}
Build this gate using AND, OR and NOT gates.

XOR can be obtained by combining two NOT gates, two AND gates and one OR gate. The logic relation for the given table is as follows:

Y=\bar{A}.B+A.\bar{B}=Y_{1}+Y_{2}\\ when\: \: Y_{1}=\bar{A}.B \: \: and\: \: Y_{2}=A.\bar{B}

Y1 cna be obtained as output of AND gate I for which one input is of A through NOT gate and another input is of B. Y2 can be obtained as output of AND gate II for  which one input is of A and other input id of B through NOT gate.

Now Y can be obtained as output from OR where Y1 and Y2 are inputs of OR gate

Thus the logic circuit of this relation is given below

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