Question

A 120 V, 60 Hz power source is connected across a 800 ohms non-inductive resistance and an unkknown capacitance in series. The voltage drop across the resistor is 102 V. (a) What is the voltage drop across the capactior? (b) what is the reactance of the capacitor?

Answer 1

Answer:

(a) 18V

(b) 861.176 ohms

Explanation:

(a) KVL:

120 = 102 + Vc => Vc = 18V

(b) current i = 102÷800 = 0.1275 A

0.1275 = 120 ÷ ( 800+Zc) => Zc = 861.176 ohms

Answer 2

Answer:

(a). The voltage drop across the capacitor is 63.21 V.

(b). The reactant of the capacitor is 495.76 Ω.

Explanation:

Given that,

Voltage = 120 V

Resistance = 800 ohm

Voltage drop = 102 V

(a) We need to calculate the voltage drop across the capacitor

Using formula of voltage

$V^2=V_{R}^2+V_{C}^2$

Put the value into the formula

$120^2=102^2+V_{C}$

$V_{C}^2=120^2-102^2$

$V_{C}=\sqrt{3996}$

$V_{C}=63.21\ V$

Let the reactant of the capacitor be $X_{c}$

We need to calculate the current

Using ohm's law

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

$I=\dfrac{102}{800}$

$I=0.1275\ A$

We need to calculate the reactant of the capacitor

Using formula of reactant of the capacitor

$V_{C}=X_{C}I$

$X_{C}=\dfrac{V_{C}}{I}$

Put the value into the formula

$X_{C}=\dfrac{63.21}{0.1275}$

$X_{C}=495.76\ \Omega$

Hence, (a). The voltage drop across the capacitor is 63.21 V.

(b). The reactant of the capacitor is 495.76 Ω.