Question

A 120V, 60Hz A.C power is connected 800 $\Omega $ non-inductive resistance and
Unknown capacitance in series. The voltage drop across the resistance is found to be
102V. Then voltage drop across capacitor is
(a) 8v
(b) 102v
(c) 63v
(d) 55v

Answer 1

Hint: Voltage drop is defined as the amount of voltage loss or dissipate that occurs
Through the all part of a circuit due to impedance (resistance of Inductor, Capacitor and
Resistance). Expressive voltage drop in a circuit can cause burn dimly, heater to hear
Poorly etc.

Complete step by step answer:
Given:
Voltage = 120v
Frequency v= 60Hz
R 800$\Omega $
Calculate capacitance c = ?
Voltage through resistance $V_R$ = 102v
Voltage across capacitor $V_C$= ?

Since R and C are connected in series
\[
\therefore \,The\,voltage\,across\,R\,is \\
{V_R} = IR \\
I = \dfrac{{{V_R}}}{R} = \dfrac{{102}}{{800}} = 0.128A \\
The\,total\,voltage\,in\,a\,circuit\,is\,given\,by \\
V = \sqrt {{V_R}^2 + {V_C}^2} \\
Squaring\,both\,side \\
{V^2} = {V_R}^2 + {V_C}^2 \\

\Rightarrow {V_C} = \sqrt {{V^2} + {V_R}^2} \\
\Rightarrow {V_C} = \sqrt {{{\left( {120} \right)}^2} + {{\left( {102} \right)}^2}} \\
\Rightarrow {V_C} = 63.2 \simeq 63V \\
\\
 \]
Hence the option (C) is correct

Note:Voltage drop under typical operating conditions can easily measure if turn on all
The electric equipment which is normally in operation and voltage at the service panel
That supplies the circuit in question. The motors with largest loads causes voltage drop
Which is often evidenced by flickering lights.