5.1 Describe the term impedance with reference to an RLC circuit - NSC Electrical Technology Electronics - Question 5 - 2017 - Paper 1

In the given phasor diagram, the voltage across the inductor (V_L = 80 V) is greater than the voltage across the capacitor (V_C = 50 V) which results in a leading reactive voltage. Therefore, this indicates that the circuit is inductive, as the current (I_T) lags behind the voltage (V_S).

Increasing the frequency will cause the inductive reactance (X_L) of the inductor to increase, resulting in a higher voltage across the inductor (V_L) as it is directly proportional to the frequency of the supply. Therefore, as the frequency rises, the voltage (V_L) across the coil will also rise.

To find the total voltage (V_T), we can use the formula:

$V_T = \sqrt{(V_R)^2 + (V_L - V_C)^2}$ Substituting the values:

$V_T = \sqrt{(110)^2 + (80 - 50)^2}$ $= \sqrt{110^2 + 30^2}$ $= \sqrt{12100 + 900}$ $= \sqrt{13000}$ $\approx 114.02 V$

To calculate the total current (I_T) in the parallel circuit, we apply:

$I_T = \sqrt{(I_R)^2 + (I_L)^2 - (I_C)^2}$ Substituting the given values:

$I_T = \sqrt{(5)^2 + (6)^2 + (4)^2}$ $= \sqrt{25 + 36 + 16}$ $= \sqrt{77}$ $\approx 8.77 A$

The phase angle (\theta) can be calculated using:

$\theta = \cos^{-1}\left(\frac{I_R}{I_T}\right)$ Substituting the values:

$\theta = \cos^{-1}\left(\frac{5}{5.39}\right)$ $\approx 21.93^\circ$

The inductive reactance (X_L) can be calculated as follows:

$X_L = \frac{V_T}{I_L}$ Substituting the given values:

$X_L = \frac{240}{6}$ $= 40 \Omega$