Question

a. A 1V peak to peak sine wave with a frequency of 1 kHz is applied across a 5 kΩ resistor. Find an expression for the current through the resistor as a function of time.

b. The same input voltage is applied across a single 2 µF capacitor, what is the current through the capacitor as a function of time?

c. The frequency of the sine wave is now increased to 5 kHz. How does this affect the current in each of the cases above?

Answer 1

(a) Given Given vo= I volt is peak value of voltage frequency, f= 10 3 42.-) W = 25X103 Resistance, R=5413r. Here v=vSmut = 0.5 Sis (2000 2t). In a pure resistire circuits, voltage and current- will be in phase. 1.0=0 0.5 Sun (2000nt) sent i= 1/2 = 5x10 9 -(104) Sis 12.000 at). A 4x3.14 2x3, 14x 10 x (b) given Capacitance, C = 2x10 ore. . Capacitive Reactance; X c = c 2 - FC =) X ci w 13 14 x 103 x 2 x 106 = 1000 = 79.62 gna capacitive circuit, voltage lags behind the current- by ano angle / in current in a capacitive circuito ů i i krs (wt + 7/2) - & Sis 2000nt + I) -) -(6.2x103) Sis (2000t + %)...A

[ for It f= 5x103 HZ = W = 27x5X103 had/s a Resistive licuit, i- & Sis (104 kt) i = (107) sm (100005t).. A In case of Capacitive liquit, X= 1 (D 2x3.148 5x103 x 2x106 - 2003.14 enfe = X = 15,92. Current i= Vo Sin (10000 at +T/) Sim (10000+ ) 15.92 i =(31,48103) Sims (1000 ont + I) Increasing frequency decreases the capacitive reactance. Hence current through capacitor increases. But it does not alter the peak value of current in resista. It remains same.