Question

A 100 Ω resistor is connected in series with a 47 µF capacitor and a source whose maximum voltage is 5 V, operating at 100.0 Hz. Find the following. • The capacitive reactance of the circuit • The impedance of the circuit • The maximum current in the circuit • The phase angle

Answer 1

Answer:

$X_c=-33.86275385\Omega$

$|Z|=105.5778675\Omega$

$I=0.04735841062A$

$\phi=20.78612878\°$

Explanation:

The electrical reactance is defined as:

$X_c=-\frac{1}{2\pi fC}$

Where:

$f=Frequency\\C=Capacitance$

So, replacing the data provided by the problem:

$X_c=\frac{1}{2\pi *100*(47*10^{-6} )} =-33.86275385\Omega$

Now, the impedance can be calculated as:

$Z=R+jX_c$

Where:

$R=Resistance\\X_c= Capacitive\hspace{3}reactance$

Replacing the data:

$Z=100-j33.86275385$

In order to find the magnitude of the impedance we can use the next equation:

$|Z|=\sqrt{(R^2)+(X_c^2)}=\sqrt{(100)^2+(-33.86275385)^2} =105.5778675\Omega$

We can use Ohm's law to find the current:

$V=I*Z\\I=\frac{V}{Z}$

Therefore the current is:

$I=\frac{5}{100-j33.86275385}=0.04485638113+0.01518960593j$

And its magnitude is:

$|I|=\sqrt{(0.04485638113)^2+(0.01518960593)^2} =0.04735841062\Omega$

Finally the phase angle of the current is given by:

$\phi=arctan(\frac{0.01518960593}{0.04485638113})=20.78612878\°$