# A sine wave voltage is applied across a capacitor. When the frequency of the voltage is decreased, what happens to the current?

Jun 23, 2015

#### Answer:

If the frequency of voltage is decreased, then current value is also $\textcolor{b l u e}{\mathrm{de} c r e a s e d}$.

#### Explanation:

If the opposition to applied voltage is increased, then current value is decreased.

Here,the opposition is due to reactance of capacitor.

${X}_{c} = \frac{1}{2 \pi f C}$

${X}_{c}$=reactance of capacitor,

$f$=frequency of applied voltage.

${X}_{c} \propto \frac{1}{f}$[since$\frac{1}{2 \pi C} =$constant]$\text{ } \textcolor{b l u e}{\left(1\right)}$

From ohm's law,

$V = {I}_{c} {X}_{c}$

${X}_{c} = \frac{V}{I} _ c$

${X}_{c} \propto \frac{1}{I} _ c$[since V is constant]$\text{ } \textcolor{b l u e}{\left(2\right)}$

From $\text{ } \textcolor{b l u e}{\left(1\right)}$&$\textcolor{b l u e}{\left(2\right)}$;

$\frac{1}{f} \propto \frac{1}{I} _ C$

$\textcolor{g r e e n}{{I}_{c} \propto f}$

From above relation,

If the frequency of voltage is decreased, then current value is also $\textcolor{b l u e}{\mathrm{de} c r e a s e d}$.