Test: Resistance & Capacitance in Series - Electrical Engineering (EE) MCQ

In an RC circuit, the applied voltage leads the current in the circuit, hence the applied voltage leads the capacitive voltage.

Form the impedance triangle, we get tanφ= Xc/R.
Hence φ=tan^-1 (Xc/R).

For a series resistance and capacitance circuit, the phase angle is always a positive value because the current will always leads the voltage.

Form the voltage triangle, we get cosφ= VR/V.
Hence φ=cos^-1VR/V.

An R-C network is a network which consists of a resistor which is connected in series to a capacitor.

At DC, the inductor acts as short circuit because the capacitive resistance is infinity. The frequency of a DC circuit is 0. The capacitive resistance=1/(2*pi*f*C). Therefore, if the frequency is 0, the capacitive resistance is infinity and it acts as an open circuit.

In an RC series circuit, the response decays with time because according to the equation, there is an exponential decrease in the response.

The particular solution of the current equation is zero. Hence, the expression for the current in an RC circuit is:
i=(V/R)e^(-t/RC).

In a series RC circuit, the initial charging current is:
I=V/R= 200/(0.5*10⁶)= 400*10^-6 A= 400 microA.

The equation for the value of current in a discharging capacitive circuit is:
I=I0*e-t/RC. From this equation, we can see that the current is exponentially decreasing since e is raised to a negative power.