Question

A 10 ohm resistor and 400 micro farad capacitor are connected in series to a 60V sinusoidal supply. The circuit current is 5A. Calculate the supply frequency and phase angle between the current and voltage.​

Answer 1

Answer:

The supply frequency and phase angle between the current and voltage are 47.7 Hz and $cos^-^1\frac{5}{6}$  respectively.

Explanation:

From the question we have,

Resistance of the resistor(R)=10Ω

The capacitance of the capacitor(C)=400μF=400×10⁻⁶F

Sinusoidal voltage supply=60V

Current in the circuit=5A

We will use the following formulas to solve the question,

$f=\frac{\omega}{2\pi }$          (1)

$cos\phi=\frac{R}{Z}$             (2)

Where,

f=the supply frequency

Ф=phase angle between the current and voltage

Z=impedance of the circuit

The impedance of the circuit is calculated as,

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

$Z=\frac{60}{5}=12\Omega$      (3)

Also,

$Z=\sqrt{R^2+X^2_C}$        (4)

$X_C$=capacitive resistance

By substituting the values in equation (4) we get;

$12=\sqrt{(10)^2+X^2_C}$

$144=100+X^2_C$

$X^2_C=44$

$X_C=2\sqrt{11} \Omega$     (5)

And $X_C$ is also calculated as,

$X_C=\frac{1}{\omega C}$

$\omega=\frac{1}{X_C C}$

$\omega=\frac{1}{2\sqrt{11}\times 400\times 10^-^6}$

$\omega=0.03\times 10^4$

$\omega=3\times 10^2 rad/s$      (6)

By substituting equation (6) in the equation (1) we get;

$f=\frac{3\times 10^2}{2\pi }=47.7Hz$     (7)

By substituting the value of R and Z in equation (2) we get;

$cos\phi=\frac{10}{12}=\frac{5}{6}$

$\phi=cos^-^1\frac{5}{6}$      (8)

Hence, the supply frequency and phase angle between the current and voltage are 47.7 Hz and $cos^-^1\frac{5}{6}$  respectively.