Question

The current through fluorescent lights are usually limited using an inductor.

1. Obtain the relation i = i_m sin(ωt – π/2)for an inductor across which an alternating emf v = v_m sin ωt is applied.

2. Why it is better to use an inductor rather than a resistor to limit the current through the fluorescent lamp?

3. When 100 V DC source is connected across a coil a current of 1 A flows through it. When 100V, 50 Hz AC source is applied across the same coil only 0.5 A flows. Calculate the resistance and inductance of the coil.

Answer 1

1. AC voltage applied to an Inductor

Consider a circuit containing an inductor of inductance ‘L’ connected to an alternating voltage. Let the applied voltage be

V = V₀ sinωt…………(1)

Due to the ow of alternating current through coil, an emf, \(L\frac{dI}{dt}\) is produced in the coil. This induced emf is equal and opposite to the applied emf (in the case of ideal inductor).

ie. \(L\frac{dI}{dt}\) = V₀ Sinωt

dI = \(\frac{V_0}{L}\) Sinωt dt

Integrating, we get

I = \(\frac{-V_0}{L_w}cosωt\)

I = \(\frac{V_0}{L_w}sin\Big(ωt-\frac{x}{2}\Big)\)   [∵-cosθ = sin(θ -\(\frac{x}{2}\))]

I = I₀  \(Sin\Big(ωt-\frac{x}{2}\Big)\)  ....(2)

Where I₀ = \(\frac{V_0}{L_ω}\)

2. No power is developed across the inductor as heat.

3. Resistance of the coil R = \(\frac{v}{I}=\frac{100}{1}\) = 100Ω.

Current through the coil when ac source is applied.

I = \(\frac{V}{\sqrt{R^2+X^2_L}}\) 

\(\sqrt{R^2+X^2_L}=\frac{V}{I}=\frac{100}{0.5}\) = 200Ω

R² + X²_L = 200² 

X_L² = 200² – 100²

X_L = 173.2Ω

L_w = 173.2

L = \(\frac{173.2}{ω}= \frac{173.2}{2\pi f}=\frac{173.2}{2\pi\times50}\) = 0.552H