Archives

In a common -base amplifier, the phase difference between the input signal voltage and output voltage is

(a) pi/2 (b) pi/4 (c) zero (c) piSolution: (c) zero In a common-base amplifier, the phase difference between the input signal voltage and output... View Article

The part of a transistor which is heavily doped to produce a large number of majority carriers is

(a) Base (b) Emitter (c) Collector (d) None of theseSolution: (b) Emitter Emitter is heavily doped to produce a large number of majority carriers.... View Article

In a common-base mode of a transistor, the collector current is 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor (β) will be

(a) 47 (b) 49 (c) 51 (d) 48Solution: (b) 49 Base current (Ib) = Ie - Ic = 5.60-5.488 = 0.112 mA Base current amplification factor (β) = Ic/Ib... View Article

A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is

(a) 500 (b) 1000 (c) 1250 (d) 50Solution: (c) 1250 We know, Resistance gain = [output resistance]/[input resistance] = 200/100 = 2 Power gain... View Article

An amplifier has a voltage gain A_V =1000. The voltage gain in dB is

(a) 30 dB (b) 60 dB (c) 3 dB (d) 20 dBSolution: (b) 60 dB Voltage gain is a measure of the ability of two part circuit to increase the power or... View Article

A change of 8 mA in the emitter current brings a change of 7.9 mA in collector current. How much change in base current is required to have the same change of 7.9 mA in collector current? Find values of α and β.

Solution: ∆IB = 8 - 7.9 = 0.1 mA α = ∆IC/∆IE = 0.99 and β = ∆IC/∆IB = 79... View Article

In BJT, maximum current flows in which of the following?

(a) in the emitter (b) in the collector (c) in the base (d) through CB junctionSolution: (a) in the emitter In a BJT , the largest current flow... View Article

In a transistor the base is

(a) a conductor with high conductivity (b) an insulator (c) an extrinsic semiconductor (d) an intrinsic semiconductorSolution: (c) an extrinsic... View Article

In a common emitter transistor amplifier, the output resistance is 500 kΩ and the current gain β = 49. If the power gain of the amplifier is 5 × 10⁶, the input resistance is

(a) 325 Ω (b) 165 Ω (c) 198 Ω (d) 240 ΩSolution: (d) 240 Ω Power gain = β2 x Ro/Ri, ….(1) where Ro is output resistance and Ri is input... View Article

The current gain of a common base transistor circuit is 0.96. On changing the emitter current by 10 mA, the change in the collector current will be

(a) 9.6 mA (b) 0.4 mA (c) 19.6 mA (d) 24 mASolution: (a) 9.6 mA Here α = 0.96 and ∆IE = 10 mA We know, Current gain for common bases transistor... View Article

In a common emitter–amplifier, the input signal is applied across

(a) Any where (b) Emitter-collector (c) Collector-base (d) Base - emitterSolution: (d) Base - emitter The input is applied to the base, and the... View Article

In a common emitter amplifier, the load resistance of the output circuit is 1000 times the resistance of the input circuit. If α= 0.98, the voltage gain is

(a) 98000 (b) 9800 (c) 49000 (d) 4900Solution: (c) 49000 We know, for transistors, β = α/(1-α) Here α = 0.98 (given) => β = 49 Now,... View Article

If α-current gain of a transistor is 0.98. What is the value of β-current gain of the transistor?

(a) 4.9 (b) 49 (c) 96 (d) 9.6 Solution: (b) 49 We know, β = α/(1-α) = 0.98/(1-0.98) = 49... View Article

The ionic bond is absent in

(a) NaCl (b) CsCl (c) LiF (d) H2O Solution: (d) H2O NaCl, CsCl and LiF contain ionic bonds between highly electromotive alkali metal... View Article

Copper has face centred cubic (fcc) lattice with interatomic spacing equal to 2.54 A. The value of lattice constant for this lattice is

(a) 3.59 A (b) 2.54 A (c) 1.26 A (d) 5.08 A Solution: (a) 3.59 A Lattice constant for (f.c.c.) = a = interatomic spacing x √2 = 3.59 A.... View Article

Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side 10 cm, 50 turns and carrying current I (Ampere) in units of μ₀I/π is

1) 250√32) 50√33) 500√34) 5√3Solution:Answer. 3tan 30 = x/dd = x/tan 30d = (5 x 10-2/{1/√3})d = 5√3 x 10-2 B= 500√3... View Article

M	T	W	T	F	S	S
			1	2	3	4
5	6	7	8	9	10	11
12	13	14	15	16	17	18
19	20	21	22	23	24	25
26	27	28	29	30	31