Physics - EMI/AC Question with Solution | TestHub

Solution:

The circuit shown is a parallel RLC circuit. In a parallel RLC circuit, the total current $I$ is the phasor sum of the individual branch currents. For a parallel RLC circuit, the total current $I$ is given by:

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

In this specific problem, the resistor current $I_R$ is not given, but the diagram implies a parallel LC circuit (or an RLC circuit where $I_R$ is negligible or zero for the purpose of this calculation, focusing on the reactive components). Assuming it's a parallel LC circuit, the total current is the difference between the inductor current $I_L$ and the capacitor current $I_C$ because they are 180 degrees out of phase.

Given:

$I_L=0.8A$

$I_C=0.6A$

The total current $I$ is the magnitude of the difference between the reactive currents:

$I=|I_L-I_C|$

$I=|0.8A-0.6A|$

$I=0.2A$

The final answer is $\boxed{0.2A}$.