Single Loop Circuits 2 3 Single NodePair Circuits

Single Loop Circuits (2. 3); Single -Node-Pair Circuits (2. 4) Dr. Holbert January 25, 2006 ECE 201 Lect-3 1

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. ECE 201 Lect-3 2

Example: Christmas Lights I 228 W 120 V 50 Bulbs Total + – 228 W ECE 201 Lect-3 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. ECE 201 Lect-3 4

I + 228 I – + 228 W 120 V 228 I – + 228 W 228 I – 228 I + … + 228 I -120 V = 0 I = 120 V/(50 228 W) = 10. 5 m. A ECE 201 Lect-3 5

Some Comments • We can solve for the voltage across each light bulb: V = IR = (10. 5 m. A)(228 W) = 2. 4 V • This circuit has one source and several resistors. The current is Source voltage/Sum of resistances (Recall that series resistances sum) ECE 201 Lect-3 6

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 ECE 201 Lect-3 7

Voltage Division Consider two resistors in series with a voltage v(t) across them: + + R 1 v 1(t) – + v(t) R 2 – v 2(t) – ECE 201 Lect-3 8

In General: Voltage Division Consider N resistors in series: Source voltage(s) are divided between the resistors in direct proportion to their resistances ECE 201 Lect-3 9

Class Examples • Learning Extension E 2. 8 • Learning Extension E 2. 9 ECE 201 Lect-3 10

Example: 2 Light Bulbs in Parallel I I 1 R 1 I 2 R 2 + V – How do we find I 1 and I 2? ECE 201 Lect-3 11

Apply KCL at the Top Node I= I 1 + I 2 Ohm’s Law: I I 1 R 1 I 2 R 2 + V – ECE 201 Lect-3 12

Solve for V Rearrange ECE 201 Lect-3 13

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 ECE 201 Lect-3 14

Now to find I 1 • This is the current divider formula. • It tells us how to divide the current through parallel resistors. ECE 201 Lect-3 15

Example: 3 Light Bulbs in Parallel I I 1 R 1 I 2 R 2 I 3 R 3 + V – How do we find I 1, I 2, and I 3? ECE 201 Lect-3 16

Apply KCL at the Top Node I= I 1 + I 2 + I 3 ECE 201 Lect-3 17

Solve for V ECE 201 Lect-3 18

Req Which is the familiar equation for parallel resistors: ECE 201 Lect-3 19

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. ECE 201 Lect-3 20

Example: More Than One Source Is 1 Is 2 I 1 R 1 I 2 R 2 + V – How do we find I 1 or I 2? ECE 201 Lect-3 21

Apply KCL at the Top Node I 1 + I 2 = Is 1 - Is 2 ECE 201 Lect-3 22

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. ECE 201 Lect-3 23

In General: Current Division Consider N resistors in parallel: Special Case (2 resistors in parallel) ECE 201 Lect-3 24

Class Examples • Learning Extension E 2. 10 • Learning Extension E 2. 11 ECE 201 Lect-3 25