A circuit is set-up by connecting L = 100 mH, C = 5 μF and R = 100 Ω in series. An alternating emf of $(150 \sqrt{2}) volt, \frac{500}{π} Hz$ is applied across this combination. Calculate the impedance of the circuit. What is the average power dissipated in (i) the resistor (ii) the capacitor (iii) the inductor and (iv) the complete circuit?

Solution

Given, a circuit with a set of components.
Inductance, L = 100 mH
Capacitance, C = 5 μF
Resistance, R = 100

Ω

Emf applied across the combination, E =

(150 \sqrt{2}) volt

Frequency of the source, f = 500/

π

Hz

Now, using the formula for impedence of the circuit, we have

Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

Inductive reactance,

X_{L} = ωL = 2 πfL = 2 π \times \frac{500}{π} \times 100 \times 10^{- 3} = 100 Ω

Capacitive reactance,

X_{C} = \frac{1}{ωC} = \frac{1}{2 πfC} = \frac{1}{2 π (\frac{500}{π}) \times 5 \times 10^{- 6}}

X_{C} = 200 Ω

and,

Z = \sqrt{(100)^{2} + (100 - 200)^{2}} = 141.4 Ω

Current flowing across the circuit,

I = \frac{V}{Z} I = \frac{150 \sqrt{2}}{100 \sqrt{2}} = 1.5 A .

Now, average power dissipated across each component is,
(i) Across resistor is,
I²R = 1.5 x 1.5 x 100
i.e., W = 225 W.

(ii) Across capacitor is zero.

(iii) Across inductor is zero.

(iv) Average power dissipated in the complete circuit is same as the power dissipated across resistor i.e., 225 W.

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