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

In the provided phasor diagram, the voltage across the inductor $V_L$ (80 V) is greater than the voltage across the capacitor $V_C$ (50 V). This indicates a leading reactive voltage, suggesting that the current $I_T$ will lag behind the voltage $V_S$. Therefore, the circuit is classified as resistive inductive.

If the frequency of the supply increases, the inductive reactance $X_L$ of the coil will also increase, as it is directly proportional to the frequency. Consequently, the voltage across the inductor $V_L$ will increase as well.

The total voltage $V_T$ in the circuit can be calculated using the formula:
$V_T = \\sqrt{(V_R)^2 + (V_L - V_C)^2}$
Substituting the values:
$V_T = \\sqrt{(110)^2 + (80 - 50)^2}$
Calculating this gives:
$V_T = \\sqrt{110^2 + 30^2} = \\sqrt{12100 + 900} = \\sqrt{13000} \approx 114.02 V$

The total current $I_T$ can be calculated using:
$I_T = \\sqrt{(I_R)^2 + (I_L - I_C)^2}$
Substituting the values:
$I_T = \\sqrt{(5)^2 + (6 - 4)^2} = \\sqrt{25 + 4} = \\sqrt{29} \approx 5.39 A$

The phase angle $\theta$ can be calculated using the relation:
$\theta = \cos^{-1}\left( \frac{I_R}{I_T} \right)$
Substituting the values:
$\theta = \cos^{-1}\left( \frac{5}{5.39} \right) \approx 21.93°$

The inductive reactance $X_L$ can be calculated using:
$X_L = \frac{V_T}{I_L} = \frac{240}{6} = 40 \Omega$