Ncert Exercises & Solutions

A coil of 0.01H inductance and 1 $Ω$ $Ω$ resistance is connected to 200 V, 50Hz AC supply. Find the impedance of the circuit and the time lag between maximum alternating voltage and current.

Hint: The time lag between the maximum alternating voltage and current depends on the angular frequency and phase angle.

Step 1 : Given, inductance L = 0.01 H

$Step 1 : Given, inductance L = 0.01 H$

resistance R = 1 Ω, voltage (V) = 200 V

$resistance R = 1 Ω, voltage (V) = 200 V$

and frequency (f) = 50 Hz .

$and frequency (f) = 50 Hz .$

Impedance of the circuit, Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{R^{2} + {(2 πfL)}^{2}}

$Impedance of the circuit, Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{R^{2} + {(2 πfL)}^{2}}$

= \sqrt{1^{2} + {(2 \times 3.14 \times 50 \times 0.01)}^{2}}

$= \sqrt{1^{2} + {(2 \times 3.14 \times 50 \times 0.01)}^{2}}$

or Z = \sqrt{10.86} = 3.3 Ω

$or Z = \sqrt{10.86} = 3.3 Ω$

Step 2 : tanϕ = \frac{ωL}{R} = \frac{2 πfL}{R} = \frac{2 \times 3.14 \times 50 \times 0.01}{1} = 3.14

$Step 2 : tanϕ = \frac{ωL}{R} = \frac{2 πfL}{R} = \frac{2 \times 3.14 \times 50 \times 0.01}{1} = 3.14$

ϕ = {tan}^{- 1} (3.14) \approx 72 °

$ϕ = \tan^{- 1} (3.14) \approx 72 °$

Phase difference, ϕ = \frac{72 \times π}{180} rad

$Phase difference, ϕ = \frac{72 \times π}{180} rad$

Step 3 : Time lag between alternating voltage and current :

$Step 3 : Time lag between alternating voltage and current :$

Δ t = \frac{ϕ}{ω} = \frac{72 π}{180 \times 2 π \times 50} = \frac{1}{250} sec

$∆ t = \frac{ϕ}{ω} = \frac{72 π}{180 \times 2 π \times 50} = \frac{1}{250} \sec$

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