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

In the given circuit, the voltage across the inductor ($V_L = 80 V$) is greater than the voltage across the capacitor ($V_C = 50 V$). This indicates that the circuit exhibits a leading reactive voltage, leading to a condition where the current $I_S$ lags behind the supply voltage $V_S$. Therefore, this circuit can be classified as resistive-inductive.

Increasing the frequency of the supply will lead to a rise in the inductive reactance ($X_L$) of the coil. Since the inductive reactance is directly proportional to the frequency, a higher frequency will result in a greater $X_L$, which in turn may cause an increase in the voltage across the inductor, $V_L$.

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:

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

The total current $I_T$ in the parallel circuit can be found using:

$I_T = \\sqrt{(I_R^2 + (I_L - I_C)^2)}$

Applying the given 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:

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

Substituting the known values:

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

The inductive reactance $X_L$ can be found using:

$X_L = \frac{V_T}{I_L}$

Substituting the known values:

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