- RL Parallel Circuit Definition: An RL parallel circuit is a setup where a resistor and an inductor are connected in parallel and share the same voltage source.
- Voltage Equality: In an RL parallel circuit, the input voltage equals the output voltage, though currents in the resistor and inductor differ.
- Impedance Calculation: The total impedance in a parallel RL circuit involves combining the resistance and inductive reactance.
- Current Calculation: Ohm’s law is used to find the currents in the resistor and inductor branches, then summing them for total current.
- Phase Angle Range: The total phase angle in a parallel RL circuit lies between 0° (pure resistive) and -90° (pure inductive).
In an ideal RL parallel circuit, a resistor and inductor share the same two nodes of a sinusoidal voltage source Vin. A voltage Vout measured across those same nodes equals input voltage by definition. The branch currents differ in magnitude and phase.
This bare parallel RL circuit therefore has no separate voltage-output node and does not by itself define a voltage filter. Filters can still use a parallel RL impedance as part of a larger divider, unlike the simple series RL circuit example.
Let us say: IT is total phasor current drawn from the voltage source.
IR is resistor current in phase with source voltage.
IL is ideal-inductor current, lagging source voltage by 90 degrees.
θ is the signed angle of IR to resultant IT, negative for a lagging current convention.
Total current IT is the phasor sum IR + IL, not the scalar sum:
With source voltage as the zero-degree reference, IR = V/R and IL = V/(jXL) = -jV/XL:
Impedance of Parallel RL Circuit
Let Z be total complex impedance in ohms.
R is ideal branch resistance in ohms.
L is ideal branch inductance in henries.
XL = ωL is inductive reactance in ohms.
For parallel resistance R and ideal inductor impedance jXL, admittance is Y = 1/R + 1/(jXL) = 1/R – j/XL. Total impedance is its reciprocal:
Multiplying numerator and denominator by the complex conjugate (R – j XL) gives rectangular impedance form:
Analysis of a Parallel RL Circuit
Given ideal resistance, inductance, frequency and RMS source voltage, analyse the RL parallel circuit with phasors:
Step 1. Calculate inductive reactance XL = 2πfL.
Step 2. Branch voltage across the inductor and resistor equals source voltage.
Step 3. Apply phasor Ohm’s law: IR = V/R and IL = V/(jXL).
Step 4. Add branch currents vectorially to get IT = IR + IL.
Step 5. The resistor current angle is 0 degrees and ideal inductor current angle is -90 degrees relative to voltage.
Step 6. Calculate total complex impedance as Z = V/IT, or use the parallel-impedance formula above.
Step 7. Calculate the signed source-current angle from the current triangle.
For a passive ideal parallel RL circuit, source-current angle lies between 0o and -90o. It approaches 0o for the pure-resistive limit and -90o for the pure-inductive limit. The impedance angle has the opposite sign and lies between 0 and +90 degrees.
