## 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 ?

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 ,

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

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

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

= 50 A

## A transformer having efficiency 90% is working on 100 V and at 2.0 kW power. If the current in the secondary coil is 5A, calculate the current in the primary coil…

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, Is = 5 A , Ep = 100 V,

(i) Ep Ip = 2 kW = 2000 W

Ip = 2000/Ep or Ip = 2000/100 = 20 A

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

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

Es Is = η × Ep Ip

Es Is = (9/10) × 2000 = 1800 W

Es = 1800 /Is = 1800/5

Es = 360 volt

## 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

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

## 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….

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

∴ XL = 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

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

## 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…

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, Ev = 20 V , R = 100 Ω

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

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

XL = 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

## An LCR circuit has L = 10 mH. R = 3 ohm and C = 1μF connected in series to a source of 15 cosωt volt. What is average power dissipated per cycle…

Q: An LCR circuit has L = 10 mH. R = 3 ohm and C = 1μF connected in series to a source of E = 15 cosωt volt. What is average power dissipated per cycle at a frequency that is 10% lower than the resonant frequency ?

Sol: Here, L = 10-2 H, R = 3 Ω, C = 10-6 F

Resonant frequency, $\large \omega_0 = \frac{1}{\sqrt{L C}}$

$\large \omega_0 = \frac{1}{\sqrt{10^{-2} \times 10^{-6}}}$

Actual frequency, ω = (90%)ω0

XL = ω L = 9 × 103 × 10-2 = 90 Ω

XC = 1/ωC = 1/(9 × 103 × 10-6) = 1000/9 ohm

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

$\large Z = \sqrt{3^2 + (\frac{1000}{9} – 90)^2 }$

Z = 21.3 ohm

Power dissipated / cycle = Ev Iv cosφ

$\large = E_v (\frac{E_v}{Z}) \frac{R}{Z} = (\frac{E_v}{Z})^2 R$

$\large = (\frac{E_0}{\sqrt{2}Z})^2 R = (\frac{15}{\sqrt{2}\times 21.3})^2 \times 3$

= 0.744 W

## In a series LCR circuit, R = 200 Ω , the voltage and the frequency of the main supply is 220 V and 50 Hz respectively . On taking out the capacitance from the circuit, the current lags behind the voltage by 30°…

Q: In a series LCR circuit, R = 200 Ω , the voltage and the frequency of the main supply is 220 V and 50 Hz respectively . On taking out the capacitance from the circuit, the current lags behind the voltage by 30°. The power dissipated in the LCR circuit is

Sol: Here, R = 200 Ω , EV = 220 V

In L-R circuit, tan30° = XL/R

In C-R circuit, tan30° = XC/R

∴ XL/R = XC/R or XL = XC

In L – C – R circuit, if φ is the phase difference between voltage and current, then

$\large tan\phi = \frac{X_L -X_C}{R}$

$\large tan\phi = \frac{0}{R} = 0$

tan φ = 0°

i.e., current and voltage are in the same phase.

∴ Average power $\large = E_v I_v cos\phi = \frac{E_v^2}{R}$

(∵ φ = 0)

$\large = \frac{220^2}{200}$

= 242 W

## An inductance of 200/π mH. A capacitance of 10^(-3)/π F and a resistance of 10 Ω are connected in series with an AC source…

Q: An inductance of (200/π ) mH. A capacitance of (10-3/π ) F and a resistance of 10 Ω are connected in series with an AC source of 220 V, 50 Hz. The phase angle of the circuit is

Sol: $\large L = \frac{200}{\pi} \times 10^{-3} = \frac{0.2}{\pi} H$

$\large C = \frac{10^{-3}}{\pi} F$

R = 10 Ω ; Ev = 220 V , f = 50 Hz

XL = ω L = 2 π f L = 2π × 50 × (0.2/π ) = 20 Ω

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

$\large X_C = \frac{\pi}{2\pi \times 50 \times 10^{-3}} = 10 ohm$

$\large tan\phi = \frac{X_L -X_C}{R} = \frac{20 -10}{10} = 1$

φ = 45°

## In a series LCR circuits, the voltage across the resistance, capacitance and inductance is 10 V each. If the capacitance is short circuited…

Q: In a series LCR circuits, the voltage across the resistance, capacitance and inductance is 10 V each. If the capacitance is short circuited then the voltage across the inductance ω i l l be

Sol: As VR = VL = VC ; R = XL = XC

Z = R ; V = IR = 10 volts

When capacitor is short circuited,

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

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

New current, $\large I’ = \frac{V}{\sqrt{2} R} = \frac{10}{\sqrt{2} R}$

Potential drop across inductance

$\large = I’ X_L = I’ R =\frac{10}{\sqrt{2} R} \times R$

= 10/√2 volt

## In a circuit L, C and R are connected in series with an alternating voltage source of frequency f …

Q: In a circuit L, C and R are connected in series with an alternating voltage source of frequency f . The current leads the voltage by 45° . The value of C is:

Sol: As current leads the voltage by 45°

$\large tan\phi = \frac{X_C – X_L}{R}$

$\large tan45^o = \frac{X_C – X_L}{R}$

$\large 1 = \frac{X_C – X_L}{R}$

$\large X_C – X_L = R$

$\large X_C = X_L + R$

$\large \frac{1}{\omega C} = \omega L + R$

$\large C = \frac{1}{\omega (\omega L + R)}$

$\large C = \frac{1}{2\pi f (2\pi f L + R)}$