A series LCR circuit is connected to an a.c. source of 220 V – 50 Hz. If the readings of voltmeters across resistor, capacitor and inductor are 65 V, 415 V and 204 V; and R = 100 $Ω$ , calculate: (i) current in the circuit; (ii) value of L, (iii) value of C and (iv) capacitance required to produce resonance with the given inductor L.

Solution

Given, a series LCR circuit.
Effective voltage, E_v = 200 V
Frequency of AC supply, f = 50 Hz
Resistance, R = 100 Ω
Voltage across resistor, V_R = 65 V
Voltage across capacitor, V_c = 415 V
Voltage across inductor, V_L = 204 V

(i) If, I_v is the current in the circuit, then

V_{R} = I_{V} \times R 65 = I_{V} \times 100

I_{V} = 0.65 A .

(ii) Using the formula,

V_{L} = I_{V} X_{L}

X_{L} = \frac{V_{L}}{I_{V}} = \frac{204}{0.65} = 313.85 Ω

X_{L} = ωL = 2 πfL = 313.85

L = \frac{313.85}{2 πf} = \frac{313.85}{2 \times 3.14 \times 50}

L = 1.0 H.

which is the required value of inductor.

(iii) Using the relation,

V_{C} = I_{V} X_{C}

X_{C} = \frac{V_{C}}{I_{V}} = \frac{415}{0.65} = 638.5 Ω X_{C} = \frac{1}{ωC} = \frac{1}{2 πfC}

Now,

C = \frac{1}{2 πf X_{C}} = \frac{1}{2 \times 3.14 \times 50 \times 638.5}

C = 4.99 \times 10^{- 6} F

(iv) Consider, C be the capacitance that would produce resonance with L = 1.0 H, then

f = \frac{1}{2 π \sqrt{LC'}}

C' = \frac{1}{4 π^{2} f^{2} L}

C' = \frac{1}{4 \times (3.14)^{2} \times (50)^{2} \times 1} = 10.1 \times 10^{- 6} F = 10.1 μF .

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