Q22 / 44 · NTA Abhyas (JEE Main) 2020 · Physics · Alternating Current

In the adjacent circuit, the instantaneous current equation is

A
$2 \sin (100 t - \frac{π}{4})$
Correct answer
B
$\sqrt{2} \sin (100 t - \frac{π}{4})$
C
$\sqrt{2} \sin (200 t - \frac{π}{4})$
D
$\sqrt{2} (100 t + \frac{π}{4})$

Solution

Let

ϕ

is the phase difference between

V

and

i

tan ϕ = \frac{X_{L}}{R} = \frac{ω L}{R} = \frac{100 \times 1}{100} =1

\Rightarrow θ = \frac{π}{4}

Since this is an

LR

circuit

V

will lead

i

∴

Instantaneous current equation will be

i_{} = i_{0} sin (100 t - ϕ)

where

ϕ = \frac{π}{4}

and

i_{0} = \frac{V_{0}}{Z}

i_{0} = \frac{V_{0}}{\sqrt{R^{2} + {(ω L)}^{2}}} = \frac{200}{\sqrt{{(100)}^{2} + {(100 \times 1)}^{2}}} = \sqrt{2}

∴ i = \sqrt{2} sin (100 t - \frac{π}{4})

Hence, the correct answer is option: 'B'.

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