Question

A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply.

(a) What is the maximum current in the coil?

(b) What is the time lag between the voltage maximum and the current maximum?

Answer 1

Inductance of the inductor, L = 0.50 H

Resistance of the resistor, R = 100 Ω

Potential of the supply voltage, V = 240 V

Frequency of the supply, v = 50 Hz

(a) Peak voltage is given as:

V₀ = `sqrt2"V"`

= `sqrt2 xx 240`

= 339.41 V

Angular frequency of the supply,

ω = 2πv

= 2π × 50

= 100π rad/s

Maximum current in the circuit is given as:

I₀ = `"V"_0/(sqrt("R"^2 + ω^2"L"^2))`

= `339.41/sqrt((100)^2 + (100π)^2 (0.50)^2)`

= 1.82 A

(b) Equation for voltage is given as:

V = V₀ cos ωt

Equation for current is given as:

I = I₀ cos (ωt − Φ)

Where,

Φ = Phase difference between voltage and current

At time, t = 0

V = V₀ (voltage is maximum)

For ωt − Φ = 0 i.e., at time, t = `phi/ω`,

I = I₀ (current is maximum)

Hence, the time lag between maximum voltage and maximum current is `phi/ω`.

Now, phase angle Φ is given by the relation,

`tan phi = (ω"L")/"R"`

= `(2π xx 50 xx 0.5)/100`

= 1.57

`phi` = 57.5° = `(57.5π)/180 "rad"`

ωt = `(57.5ω)/180`

t = `57.5/(180 xx 2π xx 50)`

= 3.19 × 10⁻³ s

= 3.2 ms

Hence, the time lag between maximum voltage and maximum current is 3.2 ms.