For the circuit given in the figure, the voltage V C (in volts) a

SOLUTION

The capacitive impedance is given by:

\(X_C=\frac{1}{ω C}\)

With ω = 5 and C = 1 μF, we can write:

\({{\rm{X}}_{\rm{C}}} = \frac{1}{{5 \times {{10}^{ - 6}}}}Ω\)

X_C = 200 kΩ 

The voltage across the capacitor will be:

\({{\rm{V}}_{\rm{C}}} = 5\angle 0 \times \frac{{\left( { - {\rm{j}}200} \right)}}{{200 - {\rm{j}}200}} \)

\(V_C= 5\angle 0 \times \frac{{ - {\rm{j}}\left( {1 + {\rm{j}}} \right)}}{2}\)

\(V_C= 2.5\sqrt 2 \angle - 45^\circ \)

In terms of π, the above can be written as:

\(V_C= 2.5\sqrt 2 \angle - 0.25{\rm{\pi }}\)

Converting the phasor representation back to the time domain representation, we can write:

\(\\ {{\rm{V}}_{\rm{c}}}\left( {\rm{t}} \right) = 2.5\sqrt 2 \sin \left( {5{\rm{t}} - 0.25{\rm{t}}} \right) \)