 # What is the current if the emf frequency is doubled? • The peak current that is flowing through a capacitor is 10.0 mA.
What will be the magnitude of the current if:

a. Frequency of the current is doubled?
b. The EMF peak voltage across the capacitor is doubled (at the original frequency)?
c. The frequency of the current is halved and the EMF peak voltage across the capacitor is doubled?

A capacitor is defined as an electronic component that can store electrical energy in the form of positive and negative electrical charges across its plates in the form of an electrostatic field. This results in creating a potential difference across the plate. Figure 1

Its ability to store the electrical charge across its plates is defined as Capacitance C of the capacitor, and its SI Unit is Farad (F).

Capacitive Reactance X_C is defined as the resistance to the flow of alternating current due to the capacitance of a capacitor. Its unit is Ohms as per the following formula:

$X_C=\dfrac{1}{2\pi fC}$

where:

$X_C=$ Capacitive reactance measured in ohms.
$f=$ AC frequency in Hertz.
$C=$ Capacitance in Farads.

Given as

$I=10.0 mA$

Considering the $Ohm’s$ $Law$ $of$ $Electricity$, Voltage is defined as follows:

$V=I\times\ X_C$

And,

$I=\dfrac{V}{X_C}$

By Substituting the value of Capacitive reactance $X_C$,

$I=\frac{V}{\dfrac{1}{2\pi fC}}=\ 2\pi\ fCV=10mA\$

Where,

$I=$ Peak Electric Current $= 10 mA$

$f=$ AC frequency in Hertz

$C=$ Capacitance in Farads.

$V=$ Peak Emf Voltage

$X_C=$ Capacitive Reactance

Now, we will explain the effect of increasing or decreasing frequency or voltage on the peak current passing through the capacitor.

$a.$ As per the above relation, peak current $I$ is directly proportional to frequency $f$.

$I\ \propto\ f\$

So, by doubling the frequency, the current is also doubled as shown below:

$I=2\pi\left(2f\right)CV=2\left(2\pi fCV\right)=2\times10mA=20mA$

$b.$ As per the above relation, peak current $I$ is directly proportional to peak voltage $V$.

$I\ \propto\ V\$

So, by doubling the peak voltage, the current is also doubled as shown below:

$I=2\pi\ fC(2V)=2\left(2\pi fCV\right)=2\times10mA=20mA$

$c.$ As per the above relation, peak current $I$ is directly proportional to frequency $f$ and peak voltage $V$.

$I\ \propto\ f\$

$I\ \propto\ V\$

So, if the frequency is halved and peak voltage is double, the current will remain the same, as shown below:

$I\ =2\pi(\frac{f}{2})C(2V)=\frac{2}{2}\left(2\pi fCV\right)=\frac{2}{2}\times10mA=10mA$

## Numerical Results

$a.$ If the frequency is doubled, peak current will also be doubled to $20.0 mA$.

$b.$ If the EMF peak voltage is doubled (at the original frequency), the peak current will also be doubled to $20.0 mA$.

$c.$ If the frequency is halved and the EMF voltage is doubled, the peak current will remain the same at $10.0 mA$.

## Example

A capacitor having capacitance of $106.1$ microfarads is connected to a $120$ $volt$, $60$ $hertz$ AC circuit. What is the amount of current flowing in the wire?

Solution:

Capacitance $C=106.1\ \mu\ F=106.1\ \times{10}^{-6}\ F$

Voltage $=120 V$

Frequency $=60 Hz$

First we will find the Capacitive reactance $X_C$

$X_C=\frac{1}{2\pi fC}=\frac{1}{2\times3.14\times(106.1\ \times{10}^{-6})\times60}=25\ ohms$

Considering Ohm’s Law,

$I=\frac{V}{X_C}=\frac{120}{25}=4.8\ Amps$

Image/Mathematical drawings are created in Geogebra.