42.MET March 2019 Q.10(B)

A coil having a resistance of 10 Ohm, and an inductance of 0.15 H is connected in series with a capacitor across a 100V, 50Hz supply. If the current and the voltage are in phase what will be the value of the current in the circuit and the voltage drop across the coil.

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Given

R = 10 ohm
L = 0.15 H
V = 100 V
f = 50 Hz

To find

1. Value of the current in the circuit
2. The voltage drop across the coil.

Solution

It is a series circuit , Voltage and current in phase. Therefore it satisfies the condition of resonance.

1. The value of current I = V / R
                                    = 100 / 10
                                    = 10 A

Reactance of inductor X = 2 x pi x f x L
                                        = 2 x 3.14 x 50 x 0.15
                                        =  47 ohm

Voltage drop across inductor = Current x Reactance
                                               = I x X
                                               = 10 x 47
                                               = 470 V
Because of resonance condition,
Voltage drop across Inductor = voltage drop across Capacitor = 470 V

Impedence of coil Z = √ (Resistance2 + Reactance2)
                                 = √  (R2 + X2)
                                 = √  (102 + 472)
                                 = 48.05 ohm

2. Voltage drop across coil  = Current x Impedence
                                        = 10 x 48.05
                                        = 480.5 V