Question

A capacitore of capacitance 10 $\mu$ F is connected to an oscillator giving an output voltage $\epsilon$ = (10 V) sin $\omega$ t. Find the peak currents in the circuit for $\omega$ = 10 $s^{-1}$, 100 $s^{-1}$, 500 $s^{-1}$, 1000 $s^{-1}$

Answer 1

Here C = $10\mu F=10\times {10}^{-6}$ F

= ${10}^{-5}$ F

E = (10 V) $sin\omega$ t,

$\omega =10s^{-1}$

(a) $i=\frac{E_0}{\frac{1}{\omega C}}=\frac{10}{\frac{1}{10}\times {10}^{-5}}$ A

= $1\times {10}^{-3}$ A

(b) $\omega =100s^{-1}$

i = $\frac{E_0}{\frac{1}{\omega C}}=\frac{10}{\frac{1}{100}\times {10}^{-6}}$

= $\frac{10}{{10}^3}=1\times {10}^{-2}$ A

= 0.01 A

(c) $\omega =500s^{-1}$

i = $\frac{E_0}{\frac{1}{\omega C}}=\frac{10}{\frac{1}{500}\times {10}^{-5}}$

= $10\times 500\times {10}^{-5}$

= $5\times {10}^2$ A = 0.05 A

(d) $\omega =1000s^{-1}$

i = $\frac{10}{\frac{1}{1000}\times {10}^{-5}}$

= $10\times 1000\times {10}^{-5}$

= ${10}^{-1}$ A = 0.1 A