Analog Electronics - Section 1 (5)

Directions to Solve

Analog Electronics - Section 1

33.

In a CE amplifier the input impedance is equal to the ratio of

A.	ac base voltage to ac base current
B.	ac base voltage to ac emitter current
C.	ac emitter voltage to ac collector current
D.	ac collector voltage to ac collector current

Answer: Option A

Explanation:

Input is applied to base with emitter grounded. The input impedance is the ratio of ac base voltage to ac base current.

34.

The output V0 in figure is

327-428-1

A.	-100 V
B.	-100 mV
C.	10 V
D.	10 mV

Answer: Option B

Explanation:

Input to non-inverting op-amp is -10 x 10^-6 x 10³ = -10 mV.

Therefore output = 363-428-1

35.

To protect the diodes in a rectifier and capacitor input filter circuit it is necessary to use

A.	surge resistor
B.	surge inductor
C.	surge capacitor
D.	both (a) and (b)

Answer: Option A

Explanation:

Resistor reduces surge current.

36.

In a BJT circuit a pnp transistor is replaced by npn transistor. To analyse the new circuit

A.	all calculations done earlier have to be repeated
B.	replace all calculated voltages by reverse values
C.	replace all calculated currents by reverse values
D.	replace all calculated voltages and currents by reverse values

Answer: Option D

Explanation:

All voltages and currents have reverse polarity.

37.

The input impedance of op-amp circuit of figure is

313-188-1

A.	120 k ohm
B.	110 k ohm
C.	infinity
D.	10 k ohm

Answer: Option D

Explanation:

Due to the presence of virtual ground at input, the resistance in the series path of input of inverting amplifier is input impedance.

38.

In figure, VEB = 0.6 V, β = 99. Then VC and IC are

316-221-1

A.	9.3 V and 1.98 mA respectively
B.	4.6 V and 1.98 mA respectively
C.	9.3 V and 0.02 mA respectively
D.	4.6 V and 0.02 mA respectively

Answer: Option A

Explanation:

361-221-1

V_C = 20 - 1.98 x 10^-3 x 5.4 x 10³ 1-sym-aa= 9.3 V.

39.

In figure The minimum and maximum load currents are

318-254-1

A.	0 and 60 mA
B.	0 and 120 mA
C.	10 mA and 60 mA
D.	10 mA and 120 mA

Answer: Option B

Explanation:

When RL = ∞, IL = 0,

When RL = 100 Ω, 362-254-1 or 120 mA.

40.

If an amplifier with gain of - 1000 and feedback factor β = - 0.1 had a gain change of 20% due to temperature, the change in gain of the feedback amplifier would be

A.	10%
B.	5%
C.	0.2%
D.	0.01%

Answer: Option C

Explanation:

367-580-1

As we know, Gain with feedback

367-580-2

A = - 1000, β = -0.1 .

Analog Electronics - Section 1 (5)

Analog Electronics - Section 1

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