A capacitor of capacitance 150.0 F is connected to an alternating source of emf given by E = 36 (120 t) V. The maximum value of current in the circuit is approximately equal to:

Given: E = 36 (120 t) V and C = 150 F. Find: The maximum current in the circuit. For a capacitor in an AC circuit, Q = CV and i = dQdt = CE t From E = 36 (120 t), we identify E0 = 36 V, = 120 rad/s Therefore, the maximum current is i0 = CE0 Substituting the values, i0 = 150 10^-6 36 120 = 2.03 A Therefore, the maximum current is approximately 2 A. The correct option is D.

A capacitor of capacitance 150.0 F is connected to an alternating… — Solution | JEE Physics

MCQEasyJEE 2023Alternating Current Basics

A capacitor of capacitance $150.0 \, \mu \text{F}$ is connected to an alternating source of emf given by $E = 36 \sin (120 \pi t)$ $\text{V}$ . The maximum value of current in the circuit is approximately equal to:

A
$\sqrt{2}$ $\text{A}$
B
$2\sqrt{2}$ $\text{A}$
C
$\frac{1}{\sqrt{2}}$ $\text{A}$
D
$2$ $\text{A}$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: $E = 36 \sin(120\pi t) \, \text{V}$ and $C = 150 \, \mu \text{F}$ .

Find: The maximum current in the circuit.

For a capacitor in an AC circuit,

Q = CV

and

i = \frac{dQ}{dt} = CE\omega \cos \omega t

From $E = 36 \sin(120\pi t)$ , we identify

E_0 = 36 \, \text{V}, \quad \omega = 120\pi \, \text{rad/s}

Therefore, the maximum current is

i_0 = CE_0\omega

Substituting the values,

i_0 = 150 \times 10^{-6} \times 36 \times 120\pi = 2.03 \, \text{A}

Therefore, the maximum current is approximately $2 \, \text{A}$ . The correct option is D.

Common mistakes

Using the resistance formula for AC circuits is incorrect here because the circuit contains only a capacitor. Use the capacitor current relation $i = C\frac{dV}{dt}$ instead.
Taking $36 \, \text{V}$ as RMS voltage is incorrect because the given expression $E = 36\sin(120\pi t)$ shows that $36 \, \text{V}$ is the maximum value. Use $E_0 = 36 \, \text{V}$ .
Missing the angular frequency $\omega = 120\pi$ is a common error. The coefficient of $t$ inside the sine function gives $\omega$ directly.

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