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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?
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Solution
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.
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