Question

A $40\mu \text{F}$ capacitor is connected to a $200\text{V}$ ac source. The RMS value of the current in the circuit is $2.5\text{A}$ then the frequency of the AC source is (nearly)

• A. $25\text{Hz}$
• B. $50\text{Hz}$
• C. $75\text{Hz}$
• D. $100\text{Hz}$

Answer 1

The correct option is B $50\text{Hz}$

$C=40\mu \text{F}=40\times {10}^{-6}\text{F};V_{rms}=200\text{V},I_{rms}=2.5\text{A}$

Capacitive reactance, $X_C=\frac{V_{rms}}{I_{rms}}=\frac{200}{2.5}=80\Omega$

$\Rightarrow \frac{1}{C\omega }=80$

$\Rightarrow \omega =\frac{1}{80\times C}=\frac{1}{80\times 40\times {10}^{-6}}=\frac{{10}^4}{32}$

$2\pi f=\frac{{10}^4}{32}$

$\therefore f=\frac{{10}^4}{2\times 3.14\times 32}=\frac{10000}{200}=50\text{Hz}$