ECE201 Lect-31 Single Loop Circuits (2.3); Single-Node-Pair Circuits (2.4) Dr. Holbert January 25, 2006

ECE201 Lect-32 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-33 Example: Christmas Lights +–+– 120V 228  50 Bulbs Total 228  I

ECE201 Lect-34 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 I + 228I + … + 228I -120V = 0 I = 120V/(50  228  ) = 10.5mA +–+– 120V 228  I+– 228I + – + –

ECE201 Lect-36 Some Comments We can solve for the voltage across each light bulb: V = IR = (10.5mA)(228  = 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-37 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-38 Voltage Division Consider two resistors in series with a voltage v(t) across them: R1R1 R2R2 – v 1 (t) + + – v 2 (t) + – v(t)

ECE201 Lect-39 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-310 Class Examples Learning Extension E2.8 Learning Extension E2.9

ECE201 Lect-311 Example: 2 Light Bulbs in Parallel How do we find I 1 and I 2 ? I R1R1 R2R2 V + – I1I1 I2I2

ECE201 Lect-312 Apply KCL at the Top Node I= I 1 + I 2 Ohm’s Law: I R1R1 R2R2 V + – I1I1 I2I2

ECE201 Lect-313 Solve for V Rearrange

ECE201 Lect-314 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-315 Now to find I 1 This is the current divider formula. It tells us how to divide the current through parallel resistors.

ECE201 Lect-316 Example: 3 Light Bulbs in Parallel How do we find I 1, I 2, and I 3 ? I R2R2 V + – R1R1 I1I1 I2I2 R3R3 I3I3

ECE201 Lect-317 Apply KCL at the Top Node I= I 1 + I 2 + I 3

ECE201 Lect-318 Solve for V

ECE201 Lect-319 R eq Which is the familiar equation for parallel resistors:

ECE201 Lect-320 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-321 Is2Is2 V R1R1 R2R2 + – I1I1 I2I2 Example: More Than One Source How do we find I 1 or I 2 ? Is1Is1

ECE201 Lect-322 Apply KCL at the Top Node I 1 + I 2 = I s1 - I s2

ECE201 Lect-323 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-324 In General: Current Division Consider N resistors in parallel: Special Case (2 resistors in parallel)

ECE201 Lect-325 Class Examples Learning Extension E2.10 Learning Extension E2.11