In the given circuit, the current drawn from the source is

20 A
10 A
5 A
5√2 A

A transformer is used to light a 140 W, 24 V bulb from a 240 V a.c. mains. The current in the main cable is 0.7 A. The efficiency of the transformer is

63.8 %
83.3 %
16.7 %
36.2 %

Solution

Power of source = EI = 240 × 0.7 = 166

⇒ Efficiency =\(\frac{140}{160}\Rightarrow \eta =83.3\)%

etermine the rms value of the emf given by

E (in volt) = 8 sin (ωt) + 6s in (2ωt)

5√2V
7√2V
10 V
10√2V

Solution

The primary winding of transformer has 500 turns where as its secondary has 5000 turns. The primary is connected to an A.C. supply of 20 V, 50Hz. The secondary will have an output of

2 V, 5 Hz
200 V,500 Hz
2V, 50 Hz
200 V,50 Hz

Solution

In a LCR circuit at resonance which of these will effect the current in circuit

R only
L and R only
R and C only
all L, C and R

Solution

At resonance, \(\omega L=\frac{1}{\omega C}\)

Hence the impedance of the circuit would be just equal to R(minimum). In other words, the LCR-series circuit will behave as a purely resistive circuit. Due to this the current is maximum. This condition is known as resonance

∴ Z = R, Current = ^V⁄_R

An alternating current is given by

i = i₁ cosωt + i₂ sinωt

The rms current is given by

\(\frac{i_{1}+i_{2}}{\sqrt{2}}\)
\(\frac{\left |i_{1}+i_{2} \right |}{\sqrt{2}}\)
\(\sqrt{\frac{i_{1}^{2}+i_{2}^{2}}{2}}\)
\(\sqrt{\frac{i_{1}^{2}+i_{2}^{2}}{\sqrt{2}}}\)

A step down transformer reduces 220 V to 110 V.The primary draws 5 ampere of current and secondary supplies 9 ampere.The efficiency of transformer is

Solution

A 12Ω resistor and a 0.21 henry inductor are connected in series to an a.c. source operating at 20 volt, 50 cycle. The phase angle between the current and source voltage is

30º
40º
80º
90º

Solution

In a RLC circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to

Solution

We know that \(f=\frac{1}{2\pi \sqrt{(LC)}}\)

when C is doubled, L should be halved so that resonant frequency remains unchanged.

The r.m.s value of an a.c. of 50 Hz is 10 amp. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be

2 × 10^–2 sec and 14.14 amp
1 × 10^–2 sec and 7.07 amp
5 × 10^–3 sec and 7.07 amp
5 × 10^–3 sec and 14.14 amp

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