Question

An alternating emf of e = 200 sin 100nt is applied across a capacitor
capacitance 2uF. The capacitive reactance and the peak current is
(A) 5x103
12,4 x 10-TA
(B) 4 x 10-27 12, 5x10A
TT
(C)
4x10-2
12,5 x 103 A
(D) 41,5 x 10-3 A
A​

Answer 1

Solution :

As per the given equation ,

• E = 200 sin100πt

Now ,

• E = E₀ sinωt

here ,

• E₀ = peak voltage
• ω = angular velocity
• t = time

On comparing both these equations we get ,

• E₀ = 200
• ω = 100π

(a) Capacitive reactance

X꜀ = 1 / ωC

here ,

• X꜀ = Capacitive reactance

• ω = angular velocity

• C = capacitance  

⇒ X꜀ = 1 / 100 x π x 2 x 10⁻⁶

⇒ X꜀ = 1 / 6.28 x 10⁻⁴

⇒ X꜀ = 0.159 x 10⁴ Ω

The value of capacitive reactance is  0.159 x 10⁴ Ω.

(2)Peak current

I₀ = E₀ / X꜀

here ,

⇒ I₀ = 200 / 0.159 x 10⁴

⇒ I₀ = 1257.8 x 10⁻⁴

⇒ I₀ = 0.126 A (approx.)

The value of peak current is 0.126 A .