Question

A capacitor of capacitance 10 μF is connected to an oscillator with output voltage ε = (10 V) sin ωt. Find the peak currents in the circuit for ω = 10 s⁻¹, 100 s⁻¹, 500 s⁻¹ and 1000 s⁻¹.

Answer 1

Capacitance of the capacitor, C = 10 μF = 10 × 10⁻⁶ F = 10⁻⁵ F
Output voltage of the oscillator, ε = (10 V)sinωt
On comparing the output voltage of the oscillator with
` ε = ε_0 `, we get:
Peak voltage ε₀ = 10 V
For a capacitive circuit,
Reactance, `X_e=1/(omegaC)`
Here, `omega` = angular frequency
           C = capacitor of capacitance
 Peak current, `I_0 = ε_0 /X_e`
(a) At ω = 10 s⁻¹:
Peak current,
I₀ = `ε_0/X_e`

= `ε_0/(1/omegaC)`

= `10/(1//10xx10^-5 )A`

      = 1 × 10⁻³ A
(b)  At ω = 100 s⁻¹:
Peak current, I₀ = `ε_0 /(1//omegaC)`
⇒` I_0 = 10/(1/100xx10^-5)`

⇒ `I_0 = 10/(1//100xx10^-5)`

⇒  `I_0 = 10/10^3 = 1xx10^-2 A`

= 0.01 A
(c) At ω = 500 s⁻¹:
Peak current, I₀ = `ε_0/(1//omegaC)`

`I_0 = epsilon_0/(1//omegaC)`

`⇒ I_0 = 10/(1//5xx10^-5)`

= `5xx10^-2 A =0.05  A`


(d) At ω = 1000 s⁻¹:
Peak current, I₀ = `epsilon_0/(1/omegaC)`

⇒ `I_0 = 10/(1//1000xx10^-5)`
⇒ `I_0 =10xx1000xx10^-5`
⇒ `I_0= 10^-1 A = 0.1 A`