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
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