Find the root mean square value of current through a capacitor of capacitance 10 μF , when connected to an a.c. source of 110 volts at 50 cycles / sec supply. What is its reactance?

The given a.c. source is V = 110 sin ω t Here, ω = 2 π f = 2 × 50 π = 100 π , substituting this in above equation We get, V = 110 sin ω t = 110 sin 100 π t Then, V r m s = 110 V 2 = 77 . 78 V So, root mean square of current: I r m s = V r m s X C = ω C V r m s = 100 π × 10 × 10 - 6 F × 77 . 78 V = 244 . 35 × 10 - 3 A = 0 . 244 A Reactance of capacitor: X C = 1 ω C = 1 100 × π × 10 - 5 F = 318 . 309 Ω Hence, the r-m-s of current through the capacitor is 0 . 244 A and the reactance of capacitor is 318 . 309 Ω .

Obtain an expression for the average power of an a.c. circuit.

In a.c. circuit, the power supplied by the source will be: P a . c . = v i = v ∘ sin ω t i ∘ sin ω t + φ P a . c . = v ∘ i ∘ 2 cos φ - cos 2 ω t + φ Here, the term cos 2 ω t + φ is time dependent and the average value is zero. Thus, P a . c . = v ∘ i ∘ 2 cos φ = v i cos φ = i 2 Z cos φ Here, Z Impedance present in the circuit. Hence, the average power of an a.c. circuit is i 2 Z cos φ .

The equation of an alternating current is I = 50 sin 100 t . Find the frequency.

Here, the alternating current is I = 50 sin 100 t . By comparing it with general equation I = I ∘ sin ω t The angular frequency: ω = 100 ⇒ 2 π f = 100 ⇒ f = 100 2 π = 50 π Hz Hence, the frequency of the given alternating current is 50 π Hz .