EEL 3004 Electrical Networks Lecture 14 SteadyState DC

EEL 3004 Electrical Networks Lecture #14 Steady-State (DC) RLC Circuits

Lecture Objectives The lecture today will focus on: Ø Introduction to steady-state (DC) analysis of RLC Circuits Ø Circuits with only DC sources & without switches. Ø Examples

RECALL Capacitor Observations • Like the resistor, the capacitor exhibits an algebraic relationship between its branch voltage and the stored charge on its plate. • Time varying voltage applied across the capacitor will induce current flow in the capacitor caused by charge variations through the charging process. • Constant voltage (DC) applied across the capacitor will result in zero capacitor current. • If the voltage across the capacitor changes its value instantly i. e. dt=0, then an infinite current is required. hence,

RECALL Inductor Observations • Like the resistor, the inductor exhibits an algebraic relationship between its terminal current and the induced magnetic flux in the core. • When time varying current is applied across the inductor, it induces voltage across the inductor winding caused by magnetic field variation. • When the inductor current is a fixed DC value, there is no voltage induced across it. • If the inductor current changes its value instantly i. e. dt=0, then the resultant induced voltage will be infinity. hence,

RLC circuits with DC sources and NO Switching (Steady-state) If the circuit consists only of direct current and voltage sources (dc), then the capacitor behaves as an open circuit, and the inductor behaves as a short circuit. Under DC conditions – the circuit has been undisturbed for a long time

Example Under DC conditions, find: a) The current passing through the inductor, b) The energy stored in the inductor Variable Values Unit VDC 2 V IDC 20 m. A R 1 1. 2 kΩ R 2 1. 6 kΩ R 3 2. 4 kΩ R 4 3. 4 kΩ L 56 m. H

The voltage across the inductor is considered zero under dc conditions:

The voltage across the inductor is considered zero under dc conditions: Simplifying the circuit yields

Where Req is given by:

Where Req is given by: Applying KCL at node V 1 yields:

Where Req is given by: Applying KCL at node V 1 yields: Solving for V 1 yields:

Now IL can be calculated by voltage division:

Now IL can be calculated by voltage division: The energy stored in the inductor is given by:

Example For the RLC circuit under DC conditions, find: a) The inductor currents and capacitor voltages, b) The energy stored in each inductor and capacitor. Variable Values Unit V 8. 6 V I 4. 6 m. A R 1 1. 2 kΩ R 2 1. 6 kΩ R 3 2. 4 kΩ L 1 120 m. H L 2 360 m. H L 3 320 m. H C 1 240 F C 2 340 F

The inductor acts as a short circuit, the capacitor like an open circuit under dc conditions:

The energy stored in the inductors is given by:

The energy stored in the capacitors is given by:

Example a) Design the load resistor RL so that in steady-state the energy stored in the capacitor and inductor are equal. b) Find the voltage VL. c) Find the power dissipated in RL. Variable Values Unit VS 6 V R 200 Ω L 250 m. H C 25 F

In steady state, the circuit becomes: The current in the inductor is given by:

In steady state, the circuit becomes: The current in the inductor is given by: Energy stored in the inductor is given by:

In steady state, the circuit becomes: The current in the inductor is given by: Energy stored in the inductor is given by: The voltage in the capacitor is given by:

In steady state, the circuit becomes: The current in the inductor is given by: The voltage in the capacitor is given by: Energy stored in the inductor is given by: Energy stored in the capacitor is given by:

In steady state, the circuit becomes: Equalizing the energy stored yields:

In steady state, the circuit becomes: Equalizing the energy stored yields: Solving for the load resistance:

In steady state, the circuit becomes: The voltage VL is easily found as:

In steady state, the circuit becomes: The voltage VL is easily found as: The power dissipated in RL is given by: