A capacitor and a resistor are connected in series with an a.c. source. If the potential difference across C, R are 120 V, 90 V respectively and if the r.m.s. current of the circuit is 3A, calculate the (i) impedance, (ii) power factor of the circuit.

Solution

Given,

Rms current, $I_{rms} = 3 A$

$Voltage across resistor, V_{R} = 90 V Voltage across capaciitor, V_{C} = 120 V$

Effective volatge, $E_{rms} = \sqrt{V_{R}^{2} + V_{C}^{2}}$

$= \sqrt{90^{2} + 120^{2}} = \sqrt{22500} = 150 V$

(i) Impedance of the circuit,
$Z = \frac{E_{rms}}{I_{rms}} = \frac{150}{3} = 50 Ω$

(ii) Power factor, $\cos ϕ = \frac{V_{R}}{E_{rms}} = \frac{90}{150} = 0.6$

For Daily Free Study Material Join wiredfaculty

Electrostatic Potential And Capacitance

A capacitor and a resistor are connected in series with an a.c. source. If the potential difference across C, R are 120 V, 90 V respectively and if the r.m.s. current of the circuit is 3A, calculate the (i) impedance, (ii) power factor of the circuit.

Some More Questions From Electrostatic Potential and Capacitance Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction