Lect3EEE 2021 Voltage and Current Division; Superposition Dr. Holbert January 23, 2008

Lect3EEE 2022 Single Loop Circuit The same current flows through each element of the circuit—the elements are in series We will consider circuits consisting of voltage sources and resistors +–+– VSVS R R R I

Lect3EEE 2023 Solve for I In terms of I, what is the voltage across each resistor? Make sure you get the polarity right! To solve for I, apply KVL around the loop +–+– VSVS R R R I+– I R + – + – N Total Resistors IR + IR + … + IR – V S = 0 I = V S / (N  R)

Lect3EEE 2024 In General: Single Loop The current i(t) is: This approach works for any single loop circuit with voltage sources and resistors Resistors in series

Lect3EEE 2025 Voltage Division Consider two resistors in series with a voltage v(t) across them: R1R1 R2R2 – v1(t)v1(t) + + – v2(t)v2(t) + – v(t)v(t)

Lect3EEE 2026 In General: Voltage Division Consider N resistors in series: Source voltage(s) are divided between the resistors in direct proportion to their resistances

Lect3EEE 2027 Two Resistors in Parallel How do we find I 1 and I 2 ? I R1R1 R2R2 V + – I1I1 I2I2

Lect3EEE 2028 Apply KCL with Ohm’s Law I R1R1 R2R2 V + – I1I1 I2I2

Lect3EEE 2029 Equivalent Resistance If we wish to replace the two parallel resistors with a single resistor whose voltage-current relationship is the same, the equivalent resistor has a value of: Definition: Parallel - the elements share the same two end nodes

Lect3EEE Now to find I 1 This is the current divider formula It tells us how to divide the current through parallel resistors

Lect3EEE Three Resistors in Parallel I= I 1 + I 2 + I 3 I R2R2 V + – R1R1 I1I1 I2I2 R3R3 I3I3

Lect3EEE Solve for V

Lect3EEE Equivalent Resistance (R eq ) Which is the familiar equation for parallel resistors:

Lect3EEE Current Divider This leads to a current divider equation for multiple parallel resistors For 2 parallel resistors, it reduces to a simple form Note this equation’s similarity to the voltage divider equation

Lect3EEE Example: More Than One Source How do we find I 1 or I 2 ? Is1Is1 Is2Is2 V R1R1 R2R2 + – I1I1 I2I2

Lect3EEE Apply KCL at the Top Node Is1Is1 Is2Is2 V R1R1 R2R2 + – I1I1 I2I2

Lect3EEE Multiple Current Sources We find an equivalent current source by algebraically summing current sources As before, we find an equivalent resistance We find V as equivalent I times equivalent R We then find any necessary currents using Ohm’s law

Lect3EEE In General: Current Division Consider N resistors in parallel: Special Case (2 resistors in parallel)

Lect3EEE Superposition “In any linear circuit containing multiple independent sources, the current or voltage at any point in the circuit may be calculated as the algebraic sum of the individual contributions of each source acting alone.”

Lect3EEE How to Apply Superposition To find the contribution due to an individual independent source, zero out the other independent sources in the circuit –Voltage source  short circuit –Current source  open circuit Solve the resulting circuit using your favorite technique(s)

Lect3EEE Superposition of Summing Circuit + – V out 1k  V1V1 V2V2 +–+– +–+– + – V ’ out 1k  V1V1 + – V ’’ out 1k  V2V2 + +–+– +–+–

Lect3EEE Use of Superposition V’ out = V 1 /3 V’’ out = V 2 /3 V out = V’ out + V’’ out = V 1 /3 + V 2 /3 + – V ’ out 1k  V1V1 + – V ’’ out 1k  V2V2 + +–+– +–+–

Lect3EEE Superposition Procedure 1.For each independent voltage and current source (repeat the following): a)Replace the other independent voltage sources with a short circuit (i.e., V = 0). b)Replace the other independent current sources with an open circuit (i.e., I = 0). Note: Dependent sources are not changed! c)Calculate the contribution of this particular voltage or current source to the desired output parameter. 2.Algebraically sum the individual contributions (current and/or voltage) from each independent source.

Lect3EEE Class Examples Drill Problems P2-2 & P2-4, P2-7, P2-1 & P2-3