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

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**Single Loop Circuits (2.3); Single-Node-Pair Circuits (2.4)**Dr. Holbert August 28, 2001 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**+**+ R1 v1(t) – v(t) + R2 v2(t) – – Voltage Division Consider two resistors in series with a voltage v(t) across them: 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**• Learning Extension E2.8 • Learning Extension E2.9 ECE201 Lect-3**Example: 2 Light Bulbs in Parallel**How do we find I1 and I2? + I1 I2 I R1 R2 V – ECE201 Lect-3**Apply KCL at the Top Node**I1 + I2 = I ECE201 Lect-3**Solve for V**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: 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**I1 + I2 + I3 = I ECE201 Lect-3**Solve for V**ECE201 Lect-3**Req**ECE201 Lect-3**No Simple Current Divider!**• We cannot make a simple current divider equation for three or more parallel resistors. • We have to solve for V, then solve for the current(s) of interest. • Definition: Parallel - the elements share the same two end nodes 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. • 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**• Learning Extension E2.10 • Learning Extension E2.11 ECE201 Lect-3