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

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Answer 1 · 2015-06-23T07:44:26

If the frequency of voltage is decreased, then current value is also $d e c r e a s e d$ $color(blue)(decreased)$ .

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

Here,the opposition is due to reactance of capacitor.

$X_{c} = \frac{1}{2 π f C}$ $X_c=1/(2pifC)$

$X_{c}$ $X_c$ =reactance of capacitor,

$f$ $f$ =frequency of applied voltage.

$X_{c} \propto \frac{1}{f}$ $X_cprop1/f$ [since $\frac{1}{2 π C} =$ $1/(2piC)=$ constant] $(1)$ $" "color(blue)((1))$

From ohm's law,

$V = I_{c} X_{c}$ $V=I_cX_c$

$X_{c} = \frac{V}{I_{c}}$ $X_c=V/I_c$

$X_{c} \propto \frac{1}{I_{c}}$ $X_cprop1/I_c$ [since V is constant] $(2)$ $" "color(blue)((2))$

From $(1)$ $" "color(blue)((1))$ & $(2)$ $color(blue)((2))$ ;

$\frac{1}{f} \propto \frac{1}{I_{C}}$ $1/fprop1/I_C$

$I_{c} \propto f$ $color(green)(I_cpropf)$

From above relation,

If the frequency of voltage is decreased, then current value is also $d e c r e a s e d$ $color(blue)(decreased)$ .