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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Physics Test1.
Calculate the range of the variable capacitor that is to be used in a tuned-collector oscillator which has a fixed inductance of 150 μH. The frequency band is from 500 kHz to 1500 kHz.
2.
The current gain of a common emitter transistor circuit shown in figure is 120. Draw the DC load line and mark the Q point on it. (VBE to be ignored).
3.
Prove the Boolean identity AC + ABC = AC and give its circuit description.
4.
In the combination of the following gates, write the Boolean equation for output Y in terms of inputs A and B.
5.
What is the output Y in the following circuit, when all the three inputs A, B, and C are first 0 and then 1?
1.
Resonant frequency, \({ f }=\frac { 1 }{ 2\pi \sqrt { LC } } \)
On simplifying, we get C = \(\frac { 1 }{ { 4\pi }^{ 2 }{ { f }^{ 2 } }L } \)
i) When frequency = 500 kHz,
\(C=\frac { 1 }{ 4\times { 3.14 }^{ 2 }\times { (500\times { 10 }^{ 3 }) }^{ 2 }\times 150\times { 10 }^{ -6 } } \)
= 676 pF
ii) When frequency = 1500 kHz,
\(C=\frac { 1 }{ 4\times { 3.14 }^{ 2 }\times { (1500\times { 10 }^{ 3 }) }^{ 2 }\times 150\times { 10 }^{ -6 } } \)
= 75 pF
Therefore, the capacitor range is 75 to 676 pF.
2.
β = 120
Base current, \({ I }_{ B }=\frac { 25V }{ 1M\Omega } =\frac { 25 }{ 1\times { 10 }^{ 6 } } =25\mu A\)
We know that
\(\beta =\frac { { I }_{ C } }{ { I }_{ B } } \) (or)
IC = β IB = 120 x 25 μA
= 3000 μA = 3 mA
VCE = VCC - ICRC
= 25 - (3 mA x 5k) = 10 V
3.
Step 1: AC (1 + B) = AC.1 [OR law-2]
Step 2: AC . 1 = AC [AND law – 2]
Therefore, AC + ABC = AC
Thus the Boolean identity is proved.
Circuit Description
4.
The output at the 1st AND gate : A\(\overline { B } \)
The output at the 2nd AND gate : ĀB
The output at the OR gate: Y = A. \(\overline { B } \) + Ā .B
5.
| A | B | C | X = A.B | Y=\(\overline { X.C } \) |
| 0 | 0 | 0 | 0 | 1 |
| 1 | 1 | 1 | 1 | 0 |
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