In the given circuit the AC source has $$\omega = 100$$ rad s$$^{-1}$$. Considering the inductor and capacitor to be ideal, what will be the current $$I$$ flowing through the circuit?

We need to determine the total rms current $$I$$ flowing through the given parallel alternating current (AC) circuit arrangement.

1. Identify the Given Parameters

From the problem, the parallel circuit consists of two parallel branches connected across an AC voltage source of $$V_{\text{rms}} = 200\text{ V}$$ with an angular frequency of $$\omega = 100\text{ rad s}^{-1}$$:

• Branch 1 (Top RC series path): Resistance $$R_1 = 100\ \Omega$$ and Capacitance $$C = 100\ \mu\text{F}$$.
• Branch 2 (Bottom RL series path): Resistance $$R_2 = 50\ \Omega$$ and Inductance $$L = 0.50\text{ H}$$.

2. Calculate Reactances and Branch Impedances

We first calculate the capacitive reactance ($$X_C$$) and inductive reactance ($$X_L$$), followed by the total impedance of each path:

• Capacitive Branch (Branch 1):
  The reactance of the ideal capacitor is:

  $$X_C = \frac{1}{\omega C} = \frac{1}{100 \times 100 \times 10^{-6}} = 100\ \Omega$$

  The impedance ($$Z_1$$) of the series RC branch is:

  $$Z_1 = \sqrt{R_1^2 + X_C^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2}\ \Omega \approx 141.42\ \Omega$$

• Inductive Branch (Branch 2):
  The reactance of the ideal inductor is:

  $$X_L = \omega L = 100 \times 0.50 = 50\ \Omega$$

  The impedance ($$Z_2$$) of the series RL branch is:

  $$Z_2 = \sqrt{R_2^2 + X_L^2} = \sqrt{50^2 + 50^2} = 50\sqrt{2}\ \Omega \approx 70.71\ \Omega$$

3. Determine Individual Branch Currents and Phase Angles

The voltage drop across both parallel branches is equal to the source voltage ($$V_{\text{rms}} = 200\text{ V}$$):

• Current in Branch 1 ($$I_1$$):
  

  $$I_1 = \frac{V_{\text{rms}}}{Z_1} = \frac{200}{100\sqrt{2}} = \sqrt{2}\text{ A} \approx 1.414\text{ A}$$

  Since it is an RC circuit, the current leads the voltage by a phase angle of $$\phi_1 = \tan^{-1}\left(\frac{X_C}{R_1}\right) = \tan^{-1}(1) = +45^\circ$$.
• Current in Branch 2 ($$I_2$$):
  

  $$I_2 = \frac{V_{\text{rms}}}{Z_2} = \frac{200}{50\sqrt{2}} = 2\sqrt{2}\text{ A} \approx 2.828\text{ A}$$

  Since it is an RL circuit, the current lags behind the voltage by a phase angle of $$\phi_2 = \tan^{-1}\left(\frac{X_L}{R_2}\right) = \tan^{-1}(1) = -45^\circ$$.

4. Calculate total Phasor Current ($$I$$)

The net phase difference between the leading branch current ($$I_1$$) and lagging branch current ($$I_2$$) is $$\Delta\phi = 45^\circ - (-45^\circ) = 90^\circ$$. Because the two current vectors are perpendicular (in quadrature), their phasor sum can be evaluated using the Pythagorean theorem:

$$I = \sqrt{I_1^2 + I_2^2}$$

$$I = \sqrt{(\sqrt{2})^2 + (2\sqrt{2})^2} = \sqrt{2 + 8} = \sqrt{10}\text{ A}$$

$$I \approx 3.16\text{ A}$$

Conclusion

The total current flowing through the AC circuit configuration is 3.16 A.