Question

Let \( i_C, i_L, \) and \( i_R \) be the currents flowing through the capacitor, inductor, and resistor, respectively, in the circuit given below. The AC admittances are given in Siemens (S). 
Which one of the following is TRUE? 

 

• A. \( i_C = 0.25 \angle 180^\circ \, A, \, i_L = 0.1 \angle 0^\circ \, A, \, i_R = 0.2 \angle 90^\circ \, A \)
• B. \( i_C = 4 \angle 180^\circ \, A, \, i_L = 10 \angle 0^\circ \, A, \, i_R = 5 \angle 90^\circ \, A \)
• C. \( i_C = 0.25 \angle 270^\circ \, A, \, i_L = 0.1 \angle 90^\circ \, A, \, i_R = 0.2 \angle 90^\circ \, A \)
• D. \( i_C = 4 \angle 90^\circ \, A, \, i_L = 10 \angle 270^\circ \, A, \, i_R = 5 \angle 0^\circ \, A \)

Hint:

To calculate AC currents, use Ohm’s law \( i = V \times Y \), where \( Y \) is the admittance and \( V \) is the voltage source.

Answer 1

The Correct Option is A

We are given the following AC admittances for the components:
Capacitor: \( Y_C = j0.25 \, S \)
Inductor: \( Y_L = -j0.1 \, S \)
Resistor: \( Y_R = 0.2 \, S \)
The voltage source is \( 1 \angle 90^\circ \, V \). To find the currents \( i_C, i_L, i_R \), we use Ohm's law for AC circuits, which states: \[ i = V \times Y. \] Step 1: Calculate the Capacitor Current
For the capacitor, the current is:
\[ i_C = V \times Y_C = 1 \angle 90^\circ \times j0.25 = 0.25 \angle 90^\circ + 90^\circ = 0.25 \angle 180^\circ \, A. \] Step 2: Calculate the Inductor Current
For the inductor, the current is: \[ i_L = V \times Y_L = 1 \angle 90^\circ \times -j0.1 = 0.1 \angle 90^\circ \, A. \] Step 3: Calculate the Resistor Current
For the resistor, the current is: \[ i_R = V \times Y_R = 1 \angle 90^\circ \times 0.2 = 0.2 \angle 90^\circ \, A. \] Thus, the currents are:
\( i_C = 0.25 \angle 180^\circ \, A \)
\( i_L = 0.1 \angle 0^\circ \, A \)
\( i_R = 0.2 \angle 90^\circ \, A \)
Therefore, the correct answer is (A).