Alternating Current

Q: A step up transformer operates on a 230 V line and a load current of 2 ampere. The ratio of the primary and secondary winding is 1 : 25. What is the current in the primary ?

Sol: using the relation ,

$\large \frac{N_s}{N_p} = \frac{I_s}{I_p} $

$\large I_s = \frac{N_s}{N_p} \times I_p $

$\large I_s = \frac{25}{1} \times 2 $

= 50 A

Q: A transformer having efficiency 90% is working on 100 V and at 2.0 kW power. If the current in the secondary coil is 5 A, calculate (i) the current in the primary coil and (ii) voltage across the secondary coil.

Sol: Here , η = 90% = 9/10, I_s = 5 A , E_p = 100 V,

(i) E_p I_p = 2 kW = 2000 W

I_p = 2000/E_p or I_p = 2000/100 = 20 A

(ii) $\large \eta = \frac{Output \; Power}{Input \; Power}$

$\large \eta = \frac{E_s I_s}{E_p I_p}$

E_s I_s = η × E_p I_p

E_s I_s = (9/10) × 2000 = 1800 W

E_s = 1800 /I_s = 1800/5

E_s = 360 volt

Q: An electric bulb has a rated power of 50 W at 100 V. If it used on an AC source of 200 V, 50 Hz, a choke has to be used in series with it. This choke should have an inductance of

Sol: Here, P = 50 W, V = 100 Volt

I = P/V = 50/100 = 0.5 A, R = V/I = 100/0.5 = 200 Ω

Let L be the inductance of the choke coil .

$\large I_v = \frac{E_v}{Z} $

$\large Z = \frac{E_v}{I_v} = \frac{200}{0.5} $

Z = 400 Ω

Now , $\large X_L = \sqrt{Z^2 – R^2}$

$\large X_L = \sqrt{400^2 – 200^2} = 200\sqrt{3} $

$\large \omega L = 200\sqrt{3} $

$\large L = \frac{200\sqrt{3}}{\omega} = \frac{200\sqrt{3}}{2 \pi f} $

$\large L = \frac{200\sqrt{3}}{100 \pi} $

L = 1.1 H

Q: An ideal choke coil takes a current of 8 ampere when connected to an AC supply of 100 volt and 50 Hz. A pure resistor under the same condition takes a current of 10 ampere. If the two are connected to an AC supply of 150 volts and 40 Hz. Then the current in a series combination of the above resistor and inductor is

Sol: For pure inductor,

$\large X_L = \frac{100}{8} = \frac{25}{2} ohm$

$\large \omega L = \frac{25}{2} ohm$

$\large L = \frac{25}{2 \omega } = \frac{25}{2 \times 2 \pi \times 50} $

= 1/8π H

$\large R = \frac{V}{I}= \frac{100}{10} $

= 10 ohm

For the combination, the supply is 150 v, 40 Hz

∴ X_L = ω L = 2π × 40 × (1/8π ) = 10 ohm

$\large Z = \sqrt{R^2 + X_L^2 }$

$\large Z = \sqrt{10^2 + 10^2 } $

= 10√2 ohm

$\large I_v = \frac{E_v}{Z} = \frac{150}{10\sqrt{2}} = \frac{15}{\sqrt{2}} A$

Q: A 750 Hertz – 20 volt source is connected to a resistance of 100 ohm, an inductance of 0.1803 henry and a capacitance of 10μf, all in series. What is the time in which the resistance (Thermal capacity = 2 joule/°C) will get heated by 10 °C?

Sol: Here, f = 750 Hz, E_v = 20 V , R = 100 Ω

L = 0.1803 H, C = 10 μF = 10_-5 F , t = ?

∆θ = 10 °C, thermal capacity = 2 J/ °C

X_L = ωL = 2 π f L = 2 × 3.14 × 750 × 0.1803

≈ 850 ohm

$\large X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C} $

$\large = \frac{1}{2 \pi \times 750 \times 10^{-5}} $

= 21.2 ohm

$\large Z = \sqrt{R^2 + (X_L – X_C)^2}$

$\large Z = \sqrt{100^2 + (850 – 21.2)^2}$

= 835 ohm

Power dissipated , $\large = E_v I_v cos\phi $

$\large = E_v (\frac{E_v}{Z}) \frac{R}{Z} = \frac{20^2 \times 100}{835}$

= 0.0574 W