A circuit containing a 80 mH inductor and a 60 μF capacitor in seriesis connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible. (a) Obtain the current amplitude and rms values. (b) Obtain the rms values of potential drops across each element. (c) What is the average power transferred to the inductor? (d) What is the average power transferred to the capacitor? (e) What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]
a)
Given: The inductance of inductor is 80 mH , the capacitance of the capacitor is 60 μF , the resistance of resistor is 15 Ω and the applied potential difference is 230 V and the frequency of signal is 50 Hz .
The peak voltage is given as,
V 0 = V 2
Where, the applied potential difference is V .
By substituting the given values in the above equation, we get
V 0 = 230 2 V = 325.22 V
The angular frequency is given as,
ω = 2 π ν
Where, the frequency of the signal is ν .
By substituting the given values in the given equation, we get
ω = 2 π ( 50 ) = 100 π rad / s
The maximum current of the circuit is given as,
I 0 = V 0 ( ω L − 1 ω C )
Where, the inductance is L , the capacitance is C and the angular frequency is ω .
By substituting the given values in the above equation, we get
I 0 = 230 2 ( 100 π × 80 × 10 − 3 − 1 100 π × 60 × 10 − 6 ) = 230 2 ( 8 π − 1000 6 π ) = − 11.63 A
Since, the value of current is negative because ω L < 1 ω C
The rms value of current is given as,
I = I 0 2
By substituting the given values in the above equation, we get
I = − 11.63 2 = − 8.22 A
Thus, the amplitude of the current is 11.63 A and the rms value of current is − 8.22 A .
b)
The potential difference across the inductor is given as,
V L = I × ω L
By substituting the given values in the above equation, we get
V L = 8.22 × 100 π × 80 × 10 − 3 = 206.61 V
The potential difference across the capacitor is given as,
V C = I × 1 ω C
By substituting the given values in the above equation, we get
V C = 8.22 × 1 100 π × 60 × 10 − 6 = 0.04363 × 10 4 = 436.3 V
Thus, the potential across the inductor is 206.61 V and the potential across the capacitor is 436.3 V .
c)
The actual voltage leads the current by angle 90 ° . So, the power factor will be zero and the value of average power consumed by the inductor will be zero.
d)
The actual voltage lags the current by angle 90 ° . So, the power factor will be zero and the value of average power consumed by the capacitor will also be zero.
e)
The average power consumed by the inductor and the average power consumed by the capacitor are zero. So, the total power absorbed over one cycle is zero.
a)
Given: The inductance of inductor is
The peak voltage is given as,
Where, the applied potential difference is
By substituting the given values in the above equation, we get
The angular frequency is given as,
Where, the frequency of the signal is
By substituting the given values in the given equation, we get
The maximum current of the circuit is given as,
Where, the inductance is
By substituting the given values in the above equation, we get
Since, the value of current is negative because
The rms value of current is given as,
By substituting the given values in the above equation, we get
Thus, the amplitude of the current is
b)
The potential difference across the inductor is given as,
By substituting the given values in the above equation, we get
The potential difference across the capacitor is given as,
By substituting the given values in the above equation, we get
Thus, the potential across the inductor is
c)
The actual voltage leads the current by angle
d)
The actual voltage lags the current by angle
e)
The average power consumed by the inductor and the average power consumed by the capacitor are zero. So, the total power absorbed over one cycle is zero.
