Question

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 ? 

• A. 6 A
• B. 4.24 A
• C. 0.94 A
• D. 5.9 A

Hint:

In parallel AC circuits, currents in different branches generally have different phase angles. Always perform phasor (vector) addition to find the total current. Scalar addition of magnitudes leads to incorrect results.

Answer 1

The Correct Option is B

Step 1: Understanding the Question: 
We are given a parallel AC circuit with two branches, one containing a resistor and a capacitor (RC branch) and the other containing a resistor and an inductor (RL branch). The objective is to determine the magnitude of the total RMS current drawn from the AC source. 
Step 2: Key Formula or Approach: 
In a parallel AC circuit, the total current is obtained by the phasor (vector) addition of the currents flowing through each branch. Since the branch currents are generally not in phase, complex number representation is used. Calculate the impedance of each branch: \(Z_1\) for the RC branch and \(Z_2\) for the RL branch. Determine the current in each branch using \(I = \frac{V}{Z}\). Add the branch currents as complex quantities to obtain the total current. Find the magnitude of the total current. 
Step 3: Detailed Explanation: 
Given: \(V_{\text{rms}} = 200\) V, \(\omega = 100\) rad/s. 
Branch 1 (Top branch): RC circuit 
Resistor: \(R_1 = 100 \, \Omega\). 
Capacitor: \(C = 100 \, \mu\text{F} = 10^{-4} \, \text{F}\). 
Capacitive reactance: \[ X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-4}} = 100 \, \Omega \] Impedance of RC branch: \[ Z_1 = R_1 - jX_C = (100 - j100)\, \Omega \] Branch 2 (Bottom branch): RL circuit 
Resistor: \(R_2 = 50 \, \Omega\). 
Inductor: \(L = 0.50 \, \text{H}\). 
Inductive reactance: \[ X_L = \omega L = 100 \times 0.50 = 50 \, \Omega \] Impedance of RL branch: \[ Z_2 = R_2 + jX_L = (50 + j50)\, \Omega \] Calculating the Currents: 
Current in RC branch: \[ I_1 = \frac{200}{100 - j100} = \frac{2}{1 - j} \] Rationalizing: \[ I_1 = \frac{2(1+j)}{(1-j)(1+j)} = \frac{2(1+j)}{2} = (1 + j1)\,\text{A} \] Current in RL branch: \[ I_2 = \frac{200}{50 + j50} = \frac{4}{1 + j} \] Rationalizing: \[ I_2 = \frac{4(1-j)}{(1+j)(1-j)} = \frac{4(1-j)}{2} = (2 - j2)\,\text{A} \] Calculating the Total Current: 
The total current is the phasor sum of branch currents: \[ I = I_1 + I_2 = (1 + j1) + (2 - j2) = (3 - j1)\,\text{A} \] Magnitude of the total current: \[ |I| = \sqrt{(3)^2 + (-1)^2} = \sqrt{10} \approx 3.16 \,\text{A} \] Analysis of the Options and Answer Key: 
The correct physical method yields a total RMS current of approximately \(3.16\,\text{A}\). However, this value may not appear directly in the given options. One of the options corresponds to a commonly made conceptual error, where the magnitudes of the branch currents are added directly instead of performing vector addition. Magnitude of current in RC branch: \[ |I_1| = \sqrt{1^2 + 1^2} = \sqrt{2}\,\text{A} \] Magnitude of current in RL branch: \[ |I_2| = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}\,\text{A} \] Incorrect scalar addition: \[ |I_1| + |I_2| = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2} \approx 4.24\,\text{A} \] This value corresponds to option (B), which arises from neglecting the phase difference between branch currents. 
Step 4: Final Answer: 
Using the correct phasor method, the magnitude of the total RMS current drawn from the source is: \[ \boxed{3.16\,\text{A}} \] Option (B) represents an incorrect scalar addition of currents and should not be used for accurate AC circuit analysis.