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A step down transformer operated on a 2.5 kV line. It supplies a load with 20 A. The ratio of the primary winding to the secondary is 10 : 1. If the transformer is 90% efficient, the voltage in the secondary is

225 V
550 V
250 V
200 V

202.

When a given coil of ohmic resistance 100 Ω and inductive reactance 100 $\sqrt{3}$ Ω is connected to a 200 V DC source, the current flows in the circuit is

1 A
1.5 A
2 A
$100 \sqrt{3}$ Ω

203.

The ratio of inductive reactance to capacitive reactance in an AC circuit will be

ω LC
ω² L
zero
ω² LC

204.

In a step up transformer, the turns ratio of field is primary and secondary is 1 : 4. A leclanche cell of emf 1.5 V is connected across the primary. The voltage developed across the secondary will be

3 V
6 V
1 V
zero

205.

When an AC source of emf e = E₀ sin (100t) is connected across a circuit, the phase difference between the emf e and the current i in the circuit is observed to be $π$ /4, as shown in the diagram. If the circuit consists possibly only of R-C or R-L or L-C is series, find the relationship between the two elements.

R = 1 kΩ , C = 10 µF
R = 1 kΩ , C = 1 µF
R = 1 kΩ , L = 10 H
R = 1 kΩ , L = 1 H

206.
For the circuit shown in figure, the voltage across 4 Ω resistance is

1 V

2 V

4 V

5 V

2 V

Let the currents in the various arms be shown in above figure.

Now using Kirchhoffs voltage law,

For closed loop ABGHA

3 I + 6 (I − I₁) − 9 = 0

or 9 I − 6 I₁ = 9

or 3 I − 2 I₁ = 3 ... (i)

For closed loop BGFGB

4 I₁ + 2 (I₁ − I₂) + 1 × I₁ − 6(I − I₁) = 0

− 6 I + 13 I₁ − 2 I₂ = 0 ... (ii)

For closed loop GDEFC

2 I₂ + 4 − 2 (I₁ − I₂) = 0

2 I₂− 4 I₂ = 4

I₁ − 2 I₂ = 2 ... (iii)

From Eq. (i) I = (3 + 2 I₁)/3

From Eq. (ii) I₂ = (I₁ − 2)/2

Putting these values in Eq. (ii), we get

$- 6 (\frac{3 + 2 I_{1}}{3}) + 13 I_{1} - \frac{2 (I_{1} - 2)}{2} = 0$

On solving I₁ = 0.5 A

Potential difference across 4 Ω = 0.5 × 4 = 2 V

207.

A LED has a voltage drop of 2V across it and passes a current of 10 mA. When it operates with a 6V battery through a limiting resistor R. The value of R is

40 kΩ
200 Ω
4 kΩ
400 Ω

208.

A series R-L-C circuit driven with E_rms = 100V and frequency f_d = 50 Hz is shown below.

What new capacitance should be connected to maximize the average power, if other parameters of the circuit are not changed ?

41.6 µF
33.5 µF
39.8 µF
31.3 µF

209.

An L-C-R series circuit with 100 Ω resistance is connected to an AC source of 200 V and angular frequency 300 rad s^-1 . When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. The average power dissipated in this L-C-R circuit is