A series RC circuit with $R=200\ \Omega$ and $C=\left(\dfrac{50}{\pi}\right)\mu\text{F}$ is connected across an a.c. source of peak voltage $\varepsilon_0=100$ V and frequency $\nu=50$ Hz. Calculate (a) the impedance of the circuit ($Z$), (b) the phase angle ($\phi$), and (c) the voltage across the resistor.

(a) $Z=200\sqrt{2}\approx282.8\ \Omega$; (b) $\phi=45^\circ$; (c) $V_R\approx70.7$ V (peak).

CBSE Class 12 · 2023 · Alternating CurrentShort Answer

A series RC circuit with

R = 200 Ω

and

C = (\frac{50}{π}) μ F

is connected across an a.c. source of peak voltage

ε_{0} = 100

V and frequency

ν = 50

Hz. Calculate
(a) the impedance of the circuit (

Z

),
(b) the phase angle (

ϕ

), and
(c) the voltage across the resistor.

Answer: (a) $Z = 2002 \approx 282.8 Ω$ ;
(b) $ϕ = 4 5^{\circ}$ ;
(c) $V_{R} \approx 70.7$ V (peak).

Capacitive reactance (the

π

cancels):

X_{C} = \frac{1}{2 π ν C} = \frac{1}{2 π \times 50 \times \frac{50}{π} \times 1 0 ^{- 6}} = \frac{1}{5 \times 1 0 ^{- 3}} = 200 Ω

(a)

Z = R^{2} + X_{C}^{2} = 20 0^{2} + 20 0^{2} = 2002 \approx 282.8 Ω

(b)

tan ϕ = \frac{X _{C}}{R} = \frac{200}{200} = 1 \Rightarrow ϕ = 4 5^{\circ}

I_{0} = \frac{ε _{0}}{Z} = \frac{100}{200 2} \approx 0.354 A

, so

V_{R} = I_{0} R = 0.354 \times 200 \approx 70.7 V

Z = 2002 Ω, ϕ = 4 5^{\circ}, V_{R} \approx 70.7 V

CBSE Class 12 Board · 2023 · Subjective