Let an a.c. circuit contain a resistance of R ohm in series with a capacitor of capacitance C farad. This circuit is shown in fig. 32(a).
Let, I = r.m.s. value of current flowing though the circuit,
VR = r.m.s. voltage across the resistance
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= IR is phase with I,
VC = r.m.s. voltage across the capacitor
= IXC lagging I by 90°,
And v = r.m.s. voltage applied across the whole circuit and is the vector sum of VR and VC.
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The vector diagram of the circuit is shown in fig. 32(c). In this diagram the current leads the applied voltage (or the voltage lags behind the current) by an angle θ, such that,
tan θ = XC/R
This is also shown in the wave diagram in fig. 32(b).
In the vector diagram VR, VC and V have formed a right-angled triangle in which VR represents the adjacent side or the base, VC represents the opposite side or the perpendicular and V represents the hypotenuse.
is the opposition to the flow of current offered jointly by the resistance and the capacitive reactance of the circuit and is called the Impedance of the circuit. It is denoted by Z and is expressed in ohm. Thus,
Impedance of the Circuit,
Current flowing through the circuit,
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I = V/Z ampere, and
Voltage applied across the circuit,
Power absorbed by the Circuit:
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Since the average power absorbed by a capacitor is nil, power absorbed by the whole circuit is equal to power absorbed by the resistance alone. Again, voltage across the resistance is VR volt and the current flowing through it is I ampere. Therefore,
power absorbed by the circuit = power absorbed by the resistance,
or P = VRI watt.
But VR = IR volt.
... P = IR x I = I2R watt.
From the vector diagram,
cos θ = VR/V
or, VR = V cos θ volt.
Hence, P = Vcos θ x I
= VI cos θ watt.
cos θ is the power factor of the circuit. Since the current leads the applied voltage, the phase angle θ is a leading angle and the power factor cos θ is a leading power factor.
From the vector diagram,
cos θ = VR/V = IR/IZ = R/Z leading.