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

The peak currents in the circuit for ω = 10 s−1 is $1 \times 10^{-3} A$

The peak currents in the circuit for 100 s−1 is $0.01 \mathrm{A}$

The peak currents in the circuit for 500 s−1 is $0.05 \mathrm{A}$

The peak currents in the circuit for 1000 s−1 is $0.1 A$

Explanation:

Condenser capacitance, C = 10 μF = $10 \times 10^{-6} \mathrm{F}=10^{-5} \mathrm{F}$

o/p oscillator voltage, $\varepsilon=(10 \mathrm{V}) \sin \omega t$

When comparing the oscillator output voltage with ε = ε0sinωt we get,

Peak to peak voltage $\varepsilon_{0}=10 V$

in case of capacitive circuit,

Reactance is , $X_{c}=\frac{1}{\omega C}$

Here, ω = angular frequency

          C = capacitor of capacitance  

Peak to peak current, $I_{0}=\frac{\varepsilon_{0}}{X_{c}}$

(a) when  ω = 10 s−1:

Peak current is ,

$I_{0}=\frac{\varepsilon_{0}}{X_{c}}$

   $=\frac{\varepsilon_{0}}{\frac{1}{w C}}$

   $=\frac{10}{\frac{1}{10 \times 10^{-5}}} A$

   $=1 \times 10^{-3} A$

(b) when  ω = 100 s−1:

    $I_{0}=\frac{\varepsilon_{0}}{\frac{1}{w C}}$

    $I_{0}=\frac{10}{\frac{1}{100 \times 10^{-5}}} A$

   $I_{0}=\frac{10}{10^{3}}$

       $=1 \times 10^{-2} A$

       = 0.01 A

(c)when ω = 500 s−1:

$I_{0}=\frac{\varepsilon_{0}}{\frac{1}{w C}}$

$I_{0}=\frac{10}{\frac{1}{500 \times 10^{-5}}} A$

$I_{0}=10 \times 500 \times 10^{-5} A$

   $=5000 \times 10^{-5} A$

   $=5 \times 10^{-2} A$

    $=5 \times 10^{-2} A$

   $=0.05 \mathrm{A}$

(d) when ω = 1000 s−1:

$I_{0}=\frac{\varepsilon_{0}}{\frac{1}{w C}}$

$I_{0}=\frac{10}{\frac{1}{1000 \times 10^{-5}}} A$

$I_{0}=10 \times 1000 \times 10^{-5} A$

   $=10000 \times 10^{-5} \mathrm{A}$

$I_{0}=10^{-1} A$

   $=0.1 A$