Single Loop Circuits (2.3); Single-Node-Pair Circuits (2.4)

Single Loop Circuits (2.3); Single-Node-Pair Circuits (2.4) Prof. Phillips January 29, 2003 ECE201 Lect-3

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. ECE201 Lect-3

Example: Christmas Lights I 228W 228W + – 50 Bulbs Total 120V 228W ECE201 Lect-3

Solve for I • The same current I flows through the source and each light bulb-how do you know this? • 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. ECE201 Lect-3

228I I + – + 228W 228I + 228I + … + 228I -120V = 0 I = 120V/(50  228W) = 10.5mA 228W 228I – + – 120V + 228W 228I – ECE201 Lect-3

Some Comments • We can solve for the voltage across each light bulb: V = IR = (10.5mA)(228W) = 2.4V • This circuit has one source and several resistors. The current is Source voltage/Sum of resistances (Recall that series resistances sum) ECE201 Lect-3

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 ECE201 Lect-3

+ + – – Voltage Division Consider two resistors in series with a voltage v(t) across them: R1 v1(t) – v(t) + R2 v2(t) ECE201 Lect-3

In General: Voltage Division Consider N resistors in series: Source voltage(s) are divided between the resistors in direct proportion to their resistances ECE201 Lect-3

Class Examples ECE201 Lect-3

+ I1 I2 I R1 R2 V – Example: 2 Light Bulbs in Parallel How do we find I1 and I2? ECE201 Lect-3

+ I1 I2 I R1 R2 V – Apply KCL at the Top Node I= I1 + I2 Ohm’s Law: ECE201 Lect-3

Solve for V Rearrange ECE201 Lect-3

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 ECE201 Lect-3

Now to find I1 • This is the current divider formula. • It tells us how to divide the current through parallel resistors. ECE201 Lect-3

Example: 3 Light Bulbs in Parallel How do we find I1, I2, and I3? + I1 I2 I3 I R1 R2 R3 V – ECE201 Lect-3

Apply KCL at the Top Node I= I1 + I2 + I3 ECE201 Lect-3

Solve for V ECE201 Lect-3

Req Which is the familiar equation for parallel resistors: ECE201 Lect-3

Current Divider • This leads to a current divider equation for three or more parallel resistors. • For 2 parallel resistors, it reduces to a simple form. • Note this equation’s similarity to the voltage divider equation. ECE201 Lect-3

Example: More Than One Source How do we find I1 or I2? + I1 I2 Is1 Is2 R1 R2 V – ECE201 Lect-3

Apply KCL at the Top Node I1 + I2 = Is1 - Is2 ECE201 Lect-3

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. ECE201 Lect-3

In General: Current Division Consider N resistors in parallel: Special Case (2 resistors in parallel) ECE201 Lect-3

Class Examples ECE201 Lect-3

Single Loop Circuits (2.3); Single-Node-Pair Circuits (2.4)