CBSE Class 12 · 2020 · Alternating CurrentShort Answer

A resistance R and a capacitor C are connected in series to a source

V = V_{0} sin ω t

. Find: (a) the peak value of the voltage across (i) the resistance and (ii) the capacitor; (b) the phase difference between the applied voltage and the current — which of them is ahead?

Solution

The impedance is

Z = R^{2} + X_{C}^{2}

with

X_{C} = \frac{1}{ω C}

, so the peak current is

I_{0} = \frac{V _{0}}{Z} = \frac{V _{0}}{R ^{2} + X _{C}^{2}}

(a)(i) Peak voltage across R:

V_{R 0} = I_{0} R = \frac{V _{0} R}{R ^{2} + X _{C}^{2}}

(ii) Peak voltage across C:

V_{C 0} = I_{0} X_{C} = \frac{V _{0} X _{C}}{R ^{2} + X _{C}^{2}}

(b) The phase angle satisfies

tan ϕ = \frac{X _{C}}{R}

. In an RC circuit the current leads the applied voltage by this angle

ϕ

CBSE Class 12 Board · 2020 · Subjective

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