In a series RC circuit with an AC source, R = 300 Ω, C = 25 μF, ε_{0} = 50 V and ν = 50/π Hz. Find the peak current and the average power dissipated in the circuit.

#### Solution

Given:

Resistance of the series RC circuit, *R* = 300 Ω

Capacitance of the series RC circuit,* C* = 25 μF

Peak value of voltage, *ε*_{0} = 50 V

Frequency of the AC source, ν = 50/ `pi` Hz

Capacitive reactance (Xc) is given by,

`X_C = 1/(omegaC)`

Here, ω = angular frequency of AC source

*C* = capacitive reactance of capacitance

∴ `X_C = 1/(2xx50/pixx25xx10^-6)`

⇒ `X_C = 10^4/25 Ω`

Net reactance of the series RC circuit`(Z) = sqrt(R^2 + (X_C)^2`

⇒ Z =`sqrt((300)^2 + (10^4/25)^2`

= `sqrt((300)^2 + (400)^2 = 500 Ω`

(a) Peak value of current `(I_0)` is given by,

`I_0 = (epsilon_0)/z`

⇒ `I_0 = 50/500 = 0.1 A`

(b) Average power dissipated in the circuit (P) is given by,

`P = epsilon_{rms}l_{rms} cosØ`

`epsilon_{rms] = epsilon_0/sqrt2`

`∴ P = E_0/sqrt2xx I_0/sqrt2xxR/Z`

⇒ P =`(50xx0.1xx300)/(2xx500)`

⇒ `P = 3/2 = 1.5 W`