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RLC Parallel Circuit Problems with Solutions

Parallel RLC Circuit Example Problems with Solutions

These questions are related to Parallel RLC Circuit which is covered in detail here:

Parallel RLC Circuit

1. What are the three characteristics of the voltage across each branch of a parallel RL circuit?
The voltage across each of the branches is the same value, equal in value to the total applied voltage, and all in phase of each other.

2. The total current in a parallel RL circuit is Equal to the vector sum rather than the arithmetic sum. Why?
Because the branch currents are out of phase with each other.

3. The terms apparent power, reactive power, and true power as they apply to the parallel RL circuit are defined as:
a. Apparent power (VA) includes both the true power (Watts) and the reactive power (VARs), the true power (WATTs) is that power dissipated by the resistive branch, and the reactive power (VARs) is the power that is returned to the source by the inductive branch.

4. The current measurements of a parallel RL circuit show a current flow of 2 amps through the resistive branch and 4 amps through the inductive branch, determine the value of the total current flow.
${{I}_{T}}=\sqrt{I_{R}^{2}+I_{L}^{2}}=4.47A$

5. For an RL circuit with a 240-V supply and 20 Ω resistor and a 48 Ω inductor calculate:
a. Apparent power.

$\begin{align}  & {{I}_{R}}=\frac{{{V}_{R}}}{R}=\frac{240}{20}=12A \\ & {{I}_{L}}=\frac{{{V}_{L}}}{{{X}_{L}}}=\frac{240}{48}=5\angle -{{90}^{o}}A \\ & \left| {{I}_{T}} \right|=\sqrt{{{\left| {{I}_{R}} \right|}^{2}}+{{\left| {{I}_{L}} \right|}^{2}}}=\sqrt{{{12}^{2}}+{{5}^{2}}}=13A \\ & S=V\left| {{I}_{T}} \right|=240*13=3120A \\\end{align}$

b. True power.

$\begin{align}  & {{I}_{R}}=\frac{{{V}_{R}}}{R}=\frac{240}{20}=12A \\ & {{P}_{R}}=V{{I}_{R}}=240*12=2880W \\\end{align}$

c. Reactive power.

$\begin{align}  & {{I}_{L}}=\frac{{{V}_{R}}}{R}=\frac{240}{48}=5\angle -{{90}^{o}}A \\ & \left| {{I}_{L}} \right|=5A \\ & {{P}_{L}}=V{{I}_{L}}=240*5=1200\text{ }VAR \\\end{align}$
d. Circuit power factor.

$\begin{align}  & \theta ={{\tan }^{-1}}\left( \frac{Q}{P} \right)={{\tan }^{-1}}\left( \frac{1200}{2880} \right)={{22.6}^{o}} \\ & Cos\theta =Cos({{22.6}^{o}})=0.923\text{ Lagging} \\\end{align}$

6. Explain the difference between RL circuit and an RC circuit.
The principal difference between two of them is the phase relationship. In a purely capacitive circuit, the current leads the voltage by 90o, while in a pure inductive the current lags the voltage by 90o.

7. Using a parallel RC circuit which has a power supply of 100 –V, 60 Hz, and a current flow through the resister of is 10 amps and the current flow through the capacitor is 10 amps. What are the following values?
a. Line current (IT).
\[{{I}_{T}}=\sqrt{I_{R}^{2}+I_{C}^{2}}=\sqrt{{{10}^{2}}+{{10}^{2}}}=14.14A\] 
b. Impedance (Z).
\[{{Z}_{T}}=\frac{{{V}_{s}}}{{{I}_{T}}}=\frac{100}{14.14}=7.07\Omega \] 
c. True power (W).
\[{{P}_{R}}=V{{I}_{R}}=100*10=1000W\] 
d. Reactive power (VARs).
\[{{Q}_{L}}=V{{I}_{L}}=100*10=1000W\]
e. Apparent power (VA).
\[{{S}_{T}}=\sqrt{P_{R}^{2}+P_{L}^{2}}=\sqrt{{{1000}^{2}}+{{1000}^{2}}}=1414.21\text{ }VA\] 
f. Power Factor

\[\begin{align}  & \theta ={{\tan }^{-1}}\left( \frac{Q}{P} \right)={{\tan }^{-1}}\left( \frac{1000}{1000} \right)={{22.6}^{o}} \\ & Cos\theta =Cos({{45}^{o}})=0.707\text{ Lagging} \\\end{align}\]

8. For a RC parallel circuit with a supply voltage of 120-V and total watt of 9604 no value for the resistor and a capacitor valued at 1500µF, determine the following:
a. The amount of current through the resistor.

\[\begin{align}  & {{P}_{R}}=\frac{V_{S}^{2}}{R}\Rightarrow R=\frac{V_{S}^{2}}{{{P}_{R}}}=\frac{{{120}^{2}}}{9604}=1.5\Omega  \\ & I=\frac{V}{R}=\frac{120}{1.5}=80A \\\end{align}\] 
b. The capacitive reactance of the capacitor.
\[{{X}_{C}}=\frac{1}{2\pi fC}=\frac{1}{2\pi *60*1500*{{10}^{-6}}}=1.77\Omega \] 
c. The amount of current flow through the capacitor.
\[{{I}_{C}}=\frac{{{V}_{s}}}{{{X}_{C}}}=\frac{120}{1.77}=67.79A\] 
d. The line current.
\[{{I}_{L}}=\sqrt{I_{R}^{2}+I_{C}^{2}}=\sqrt{{{80}^{2}}+{{67.79}^{2}}}=104.85A\] 
e. Apparent power.
\[S=V{{I}_{L}}=120*104.85=12582\text{ }VA\] 
f. PF

\[\begin{align}  & \theta ={{\tan }^{-1}}\left( \frac{R}{{{X}_{C}}} \right)={{\tan }^{-1}}\left( \frac{1.5}{1.77} \right)={{40.28}^{o}} \\ & Cos\theta =Cos({{40.28}^{o}})=0.76\text{ Leading} \\\end{align}\] 

9. 4-A inductor and a 12-A capacitor in parallel, what is the total current?

\[\begin{align}  & {{I}_{L}}=\text{ }4A\text{ } \\ & {{I}_{C}}=\text{ }12A \\ & {{I}_{C}}-\text{ }{{I}_{L}}=\text{ }{{I}_{T}}\text{= }8A \\\end{align}\]

10. With a source voltage of 208-V, frequency of 60 Hz and a LC parallel circuit that has XC = 16 Ω and XL = 8 Ω, what would be:
a. Impedance value be.
\[{{Z}_{T}}=\frac{{{Z}_{C}}*{{Z}_{L}}}{{{Z}_{C}}-{{Z}_{L}}}=\frac{16*8}{16-8}=16\Omega \]
b. Current through the capacitor.
\[{{I}_{C}}=\frac{Vs}{{{X}_{C}}}=\frac{208}{16}=13A\] 
c. Current through the inductor.
\[{{I}_{L}}=\frac{Vs}{{{X}_{L}}}=\frac{208}{8}=26A\] 
d. Line current.
\[{{I}_{T}}=\text{ }{{I}_{L}}-\text{ }{{I}_{C}}\text{= }13A\]
e. The impedance (Z).
\[{{Z}_{T}}\text{ }=\frac{{{V}_{s}}}{{{I}_{T}}}=16A\]

11. You have a parallel RLC circuit with 6-A trough the resistor, 8-A through the inductor, 5-A through the capacitor, calculate total line current. Is the circuit capacitive or inductive?
${{I}_{L}}=\sqrt{I_{R}^{2}+{{\left( {{I}_{L}}-{{I}_{C}} \right)}^{2}}}=\sqrt{{{6}^{2}}+{{3}^{2}}}=6.7A$
Since more current is through an inductor so circuit is inductive in nature.

12. You have a parallel RLC circuit with a 16 Ω resistor, 8 Ω inductor, 20 Ω capacitor, and a 120-V power supply what are the following values?
a. Current through the resistor (IR).
\[{{I}_{R}}=\frac{{{V}_{s}}}{R}=\frac{120}{16}=7.5A\] 
b. Current through the inductor (IL).
\[{{I}_{L}}=\frac{{{V}_{s}}}{{{X}_{L}}}=\frac{120}{8}=15A\]
c. Current through the capacitor (IC).
\[{{I}_{C}}=\frac{{{V}_{s}}}{{{X}_{C}}}=\frac{120}{20}=6A\]
d. Net reactive current (IX).
\[{{I}_{X}}={{I}_{L}}-{{I}_{C}}=9\text{ }A\]
e. Total line current (IT).
\[{{I}_{T}}=\sqrt{I_{R}^{2}+{{\left( {{I}_{L}}-{{I}_{C}} \right)}^{2}}}=\sqrt{{{7.5}^{2}}+{{9}^{2}}}=11.71A\]

About Ahmad Faizan

Mr. Ahmed Faizan Sheikh, M.Sc. (USA), Research Fellow (USA), a member of IEEE & CIGRE, is a Fulbright Alumnus and earned his Master’s Degree in Electrical and Power Engineering from Kansas State University, USA.

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