Question

A resistor of $40\Omega $, an inductor of $\dfrac{5}{\pi }H$ and a capacitor of $\dfrac{{50}}{\pi }\mu F$ are connected in a series across a source of alternating voltage of $140\sin 100\pi t$ Volts. Find the rms voltage across the resistor, the inductor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox.

Answer 1

Hint: Compare the given voltage with the general expression of voltage. Find the reactance of all components and use it to find V.

Formula used:
Voltage V:
$V = {V_0}\sin \omega t$ ……(1)
where,
${V_0}$is the peak voltage
$\omega $ is the angular frequency
t is the time
r.m.s voltage:
${V_{rms}} = \dfrac{{{V_0}}}{{\sqrt 2 }}$ ……(2)
Inductive reactance:
${X_L} = \omega L$ ……(3)
where,
L is the inductance
Capacitive reactance:
${X_C} = \dfrac{1}{{\omega C}}$ ……(4)
where,
C is the capacitance
Impedance:
$Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} $ ……(5)
where,
R is the resistance
Voltage:
$V = IX$ ……(6)
where,
X is the reactance
I is the current
rms Current:
${I_{rms}} = \dfrac{{{V_{rms}}}}{Z}$ ……(7)

Step-by-step answer:
Given:
1. Resistance (R) =$40\Omega $
2. Inductance (L) = $\dfrac{5}{\pi }H$
3. Capacitance (C) = $\dfrac{{50}}{\pi }\mu F$
4. Alternating voltage (V) =$140\sin 100\pi t$ Volts

To find:
The rms voltage across the resistor, the inductor and the capacitor.
Algebraic sum of these voltages.

Step 1 of 5:
Compare the given expression for alternating voltage with eq (1):
$140\sin 100\pi t = {V_0}\sin \omega t$
Comparing both sides, we get:
$
  {V_0} = 140Volts \\
  \omega = 100\pi \\
 $
Use eq (2) to find ${V_{rms}}$:
$
  {V_{rms}} = \dfrac{{140}}{{\sqrt 2 }} \\
  {V_{rms}} = \dfrac{{140}}{{1.4}} \\
  {V_{rms}} = 100V \\
 $

Step 2 of 5:
Find inductive reactance using eq (3):
${X_L} = 100\pi \times \dfrac{5}{\pi }$
${X_L} = 500\Omega $
Find capacitive reactance using eq (4):
$
  {X_C} = \dfrac{1}{{100\pi \times \dfrac{{50}}{\pi } \times {{10}^{ - 6}}}} \\
  {X_C} = 200\Omega \\
 $

Step 3 of 5:
Find the impedance using eq (5):
$
  Z = \sqrt {{{400}^2} + {{(500 - 200)}^2}} \\
  Z = 500\Omega \\
 $

Step 4 of 5:
Find the rms current in the circuit using eq (7):
$
  {I_{rms}} = \dfrac{{100}}{{500}} \\
  {I_{rms}} = 0.2A \\
 $
Find voltage across resistor using eq (6):
$
  {V_R} = 0.2 \times 400 \\
  {V_R} = 80V \\
 $
Find voltage across capacitor using eq (6):
$
  {V_C} = 0.2 \times 200 \\
  {V_C} = 40V \\
 $
Find voltage across inductor using eq (6):
$
  {V_L} = 0.2 \times 500 \\
  {V_L} = 100V \\
 $

Step 5 of 5:
Find the algebraic sum of voltages:
$V = {V_R} + ({V_L} - {V_C})$
(${V_C}$ was subtract as its phase is opposite of ${V_L}$)
$
  V = 80 + 100 - 40 \\
  V = 160V \\
 $
This is not equal to ${V_{rms}}$. The reason for this is that: ${V_R}$ lags $({V_L} - {V_C})$ by an angle of $\pi /2$ .
The correct way to sum these is:
${V_{rms}} = \sqrt {{V_R}^2 + {{({V_L} - {V_C})}^2}} $
$
  {V_{rms}} = \sqrt {{{80}^2} + {{(100 - 40)}^2}} \\
  {V_{rms}} = 100V \\
 $
Hence, the paradox is resolved.

Note: In questions like these, compare with the standard expression for Voltage. Go step by step finding the reactance and corresponding voltages.

Final answer:
R.M.S. voltages across resistor, capacitor and inductor are 80V, 40V and 100V respectively.
Yes, the algebraic sum of these voltages is greater than the r.m.s. value of voltage as ${V_R}$ lags $({V_L} - {V_C})$ by an angle of $\pi /2$.