## Parallel RLC Circuit

### Task number: 1787

A resistor, an ideal capacitor and an ideal inductor are connected in parallel to a source of alternating voltage of 160 V at a frequency of 250 Hz. A current of 2 A flows through the resistor and a current of 0.8 A flows through the inductor. The total current through the circuit is 2.5 A. Assess the resistance of the resistor, the capacity of the ideal capacitor and the inductance of the ideal inductor (presume that *I*_{C} > *I*_{L}).

Note: The assigned values of voltage and currents are the effective values.

#### Hint — circuit diagram

#### Hint — phasor diagram

A phasor diagram for a parallel alternating current circuit is drawn analogically to that for a series circuit. We must take into account that in a parallel circuit, the

*voltage*is the same across all elements, in contrast to a series circuit, where the same current flows through all elements.**How to draw the phasor diagram of a parallel**: Draw the phasor of voltage along the*RLC*circuit*x*axis as well as the phasor of current through the resistor. Draw the phasor of the current through the capacitor along the positive direction of the*y*axis and the phasor of the current through the inductor along the negative direction of the*y*axis.#### Notation

*V*= 160 VSource voltage *f*= 250 HzFrequency of the source voltage *I*_{R}= 2 ACurrent through the resistor *I*_{L}= 0.8 ACurrent through the inductor *I*= 2.5 ATotal current through the circuit *R*= ? [Ω]Resistance of the resistor *C*= ? [F]Capacity of the capacitor *L*= ? [H]Inductance of the inductor #### Analysis

We find the unknown quantities using Ohm’s law for alternating current circuits that we apply to each element. The voltage across all elements is equal to the source voltage because we deal with a parallel circuit.

We find the unknown current through the capacitor using the assigned currents and the phasor diagram for the parallel RLC circuit.

#### Solution and numerical substitution

**Resistance**:*R*of the resistorThe circuit is parallel and the voltage across the resistor

\[ V = I_\mathrm R R, \]*V*_{R}is thus the same as the source voltage*V*. Thence, we can write Ohm’s law for the alternating current circuit in the form:where

\[ R = \frac{V}{I_\mathrm R}. \]*I*_{R}is the effective value of current flowing through the resistor. We express the resistance of the resistor*R*from the equation:And we use the assigned values:

\[ R = \frac{160}{2}\,\mathrm \Omega = 80\,\mathrm \Omega. \]

**Inductance***L*of the inductor:The voltage across the inductor

\[ V = X_\mathrm L I_\mathrm L = 2 \pi f L\, I_\mathrm L, \]*V*_{L}is again the same as the source voltage*V*and we can express it from Ohm’s law for the alternating current circuit:where

*X*_{L}is the inductive reactance of the inductor,*f*is the source voltage frequency,*L*is the inductance of the inductor and*I*_{L}is the effective value of current flowing through the inductor.We express the inductance

\[ L = \frac{V}{2 \pi f I_\mathrm L}, \]*L*:and we use the assigned values:

\[ L = \frac{160}{2 \cdot \pi \cdot 250 {\cdot} 0.8}\,\mathrm H \,\dot=\, 0.13 \,\mathrm H. \]

**Capacity**:*C*of the capacitorWe derive the capacity of the capacitor

\[ V = X_\mathrm C I_\mathrm C =\frac{I_\mathrm C}{2 \pi f C},\]*C*also from Ohm’s law for the alternating current circuit:where

*X*_{C}is the capacitive reactance of the capacitor,*f*is the source voltage frequency, and*I*_{C}is the effective current through the capacitor. The voltage across the capacitor is the same as the source voltage*V*.We express the capacity of the capacitor

\[ C =\frac{I_\mathrm C}{2 \pi f V}.\tag{1}\]*C*:We draw the phasor diagram for currents in order to calculate the effective current

*I*_{C}flowing through the capacitor, taking into account that the assignment states that*I*_{C}>*I*_{L}:We express the relation between the currents from the phasor diagram:

\[ I^2 = (I_\mathrm C - I_\mathrm L)^2 + I_\mathrm R^2. \]We multiply out and rearrange:

\[ I^2 = I_\mathrm C^2 - 2I_\mathrm L I_\mathrm C +I_\mathrm L^2 + I_\mathrm R^2, \] \[ 0 = I_\mathrm C^2 - 2I_\mathrm L I_\mathrm C +(I_\mathrm L^2 + I_\mathrm R^2 - I^2).\]We solve the quadratic equation for the unknown current

\[(I_\mathrm C)_\mathrm {1{,}2} = \frac{ 2 I_\mathrm L \pm \sqrt{ (2 I_\mathrm L)^2 - 4 (I_\mathrm L^2 + I_\mathrm R^2 -I^2)}}{2}. \]*I*_{C}:We rearrange:

\[(I_\mathrm C)_\mathrm {1{,}2} = I_\mathrm L \pm \sqrt{ ( I_\mathrm L)^2 - (I_\mathrm L^2 + I_\mathrm R^2 -I^2)}, \] \[(I_\mathrm C)_\mathrm {1{,}2} = I_\mathrm L \pm \sqrt{ I^2 - I_\mathrm R^2 }. \]And we use the assigned values:

\[(I_\mathrm C)_\mathrm {1} = 0.8 + \sqrt{ 2.5^2 - 2^2 }\,\mathrm A= 2.3 \,\mathrm A ,\] \[(I_\mathrm C)_\mathrm {2} = 0.8 - \sqrt{ 2.5^2 - 2^2 }\,\mathrm A= -0.7\,\mathrm A .\]Only the positive value of the effective current

*I*_{C}= 2.3 A is physically meaningful in our case.We substitute this result into the equation for the capacity of the capacitor (1) that we derived above from Ohm’s law:

\[ C =\frac{I_\mathrm C}{2 \pi f U}= \frac{2.3}{2 \cdot \pi \cdot 250 {\cdot} 160}\,\mathrm F \,\dot=\, 9.2 {\cdot} 10^{-6} \,\mathrm F = 9.2 \,\mathrm{ \mu F}. \]#### Answer

The resistor has a resistance of 80 Ω, the inductor has an inductance of approximately 0.13 H and the capacity of the capacitor is about 9.2 μF.