Ncert Exercises & Solutions

7.13 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?

Given,

Inductance, 𝐿𝐿 = 0.50 H

Resistance, 𝑅𝑅 = 100 Ω

rms voltage, $𝑉_{rms}$ = 240 V

frequency, 𝑓 = 50 Hz

(a) The rms current in the coil,

$I_{r m s} = \frac{V_{r m s}}{\sqrt{X_{L}^{2} + R^{2}}}$

For an inductive circuit,

$\begin{array}{l} X_{L} = 2 π \times 50 H z \times 0.5 H = 157 Ω \\ I_{r m s} = \frac{240 V}{\sqrt{(157 Ω)^{2} + (100 Ω)^{2}}} = \frac{240 V}{186.1 Ω} = 1.2 A \end{array}$

So, The maximum current in the coil

$I_{m} = \sqrt{2} \times I_{r m s} = \sqrt{2} \times 1.2 A = 1.82 A$

(b) In LR circuit,

\begin{array}{l} If, V = V_{m} \cos ω t then, \\ I = I_{m} \cos (ω t - ϕ) \end{array}

Voltage is maximum at t = 0 whereas, the current is maximum at = 𝜙/𝜔,

Therefore, the time lag between the maximum voltage and maximum current = 𝜙/𝜔.

$\begin{array}{l} Also, \tan ϕ = \frac{ω L}{R} = \frac{2 π \times 50 H z \times 0.5 H}{100 Ω} = 1.57 \\ ϕ = \tan^{- 1} 1.57 = {57.5}^{\circ} = \frac{57.5 π}{180} \\ Timelag = \frac{ϕ}{ω} = \frac{57.5 π}{180 \times 2 π \times 50} = 3.2 m s \end{array}$

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