Question

A 120 V, 60 Hz power source is connected in series across an 800 Ω resistance and an unknown capacitance in series. The voltage drop across the resistor is 102 V. Find (a) the voltage drop across the capacitor; and (b) the reactance of the capacitor.

Answer 1

Answer:

a)Vc= 63.21 V

b)Xc= 495.756  Ω

Explanation:

Assume that voltage across the resistance is Vr and capacitor Vc

So

$V^2=V^2c+V^2r$

Now by putting the values

$120^2=V^2c+102^2$

Vc= 63.21 V

We know that for resistance

Vr= I R

102 = I x 800

I=0.1275 A

Let reactance of capacitor is Xc

So

Vc= Xc . I

63.21 = 0.1275 x Xc

Xc= 495.756  Ω

Answer 2

Answer:

(a) 63.21 volt

(b) 495.76 ohm            

Explanation:

We have given Voltage V = 120 Volt

Frequency f = 60 Hz

Resistance R = 800 ohm

Voltage drop across resistor $V_R=102volt$

(A) We know that resultant voltage for series circuit is given by $V=\sqrt{V_R^2+V_C^2}$

$120=\sqrt{102^2+V_C^2}$

Squaring both side $14400=10404+V_C^2$

$V_C=63.2139Volt$

(b) Current through the circuit $I=\frac{V_R}{R}=\frac{102}{800}=0.1257A$

As $V=IX_C$

So $63.21=0.1257\times X_c$

$X_c=495.76ohm$