Single Loop Circuits * with a current source * with a voltage source * with multiple sources * voltage divider circuits * Equivalent resistance Superposition method * Principle * Procedures * How to apply Lecture 3. Single Loop Circuits & Superposition Method 1

2 Single Loop Circuits The same current flows through each element of the circuit—the elements are in series. +–+– VSVS R1R1 R2R2 RnRn I V1V1 V2V2 ISIS R1R1 R2R2 RnRn V1V1 I V2V2 V3V3 V3V3 With an independent voltage sourceWith an independent current source

3 What is I? Single Loop Circuits – with a Current Source In terms of I, what is the voltage across each resistor? … ISIS R1R1 R2R2 RnRn V1V1 I V2V2 V3V3

4 To solve for I, apply KVL around the loop. +–+– VSVS R1R1 R2R2 RnRn I+– I R 2 + – I R 1 I R n + – IR 1 + IR 2 + … + IR n – V S = 0 Single Loop Circuits – with a Voltage Source In terms of I, what is the voltage across each resistor? …

5 With Multiple Voltage Sources The current i(t) is: Resistors in series

6 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) If n resistors in series:

7 Voltage Divider: A Practical Example Electrochemical Fabrication of Quantum Point Contact or Atomic-scale wire Molecular Junction

Anode: Etching delocalized, but Cathode: Deposition localized at sharpest point, due to: Self-focusing – directional growth Decreasing Gap! E Voltage Divider: An Example

Initially, R gap >> R ext, V gap ~ V 0 full speed deposition. Finally, R gap << R ext, V gap ~ 0 deposition terminates. The gap resistance is determined by R ext. Voltage Divider: An Example

2 Growth starts after applying 1.5 V 1 Two electrodes with 10  m initial separation 3 Self-terminates after forming a tunneling gap Voltage Divider: An Example

Ohmic behavior Time (sec.) G (2e 2 /h) Stepwise increase in Conductance Voltage Divider: An Example

12 Example: Two Resistors in Parallel How do you find I 1 and I 2 ? I R1R1 R2R2 V + – I1I1 I2I2

13 I R1R1 R2R2 V + – I1I1 I2I2 Apply KCL with Ohm’s Law Example: Two Resistors in Parallel

Equivalent Resistance of Parallel Resistors Two parallel resistors is often equivalent to a single resistor with resistance value of: n-Resistors in parallel:

15 What are I 1 and I 2 ? This is the current divider formula It tells us how to divide the current through parallel resistors

16 Circuits with More Than One Source How do we find I 1 or I 2 ? Is1Is1 Is2Is2 V R1R1 R2R2 + – I1I1 I2I2

17 Is1Is1 Is2Is2 V R1R1 R2R2 + – I1I1 I2I2 Apply KCL at the Top Node What if More Than One Source?

18 Class Examples Example: P1-33 (page 43). Drill Problem P1-34 (page 43).

19 Superposition Method – A More General Approach to Multiple Sources “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.”

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)

21 Superposition of Summing Circuit + – V ou t 1k  V1V1 V2V2 +–+– +–+– + – V ’ ou t 1k  V1V1 + – V ’’ ou t 1k  V2V2 + +–+– +–+–

22 V’ out = V 1 /3 V’’ out = V 2 /3 V out = V’ out + V’’ out = V 1 /3 + V 2 /3 + – V ’ ou t 1k  V1V1 + – V ’’ out 1k  V2V2 + +–+– +–+– Superposition of Summing Circuit (cont’d)

23 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.

24 Class Examples Example 2-9 (page 70). Drill Problem 2.7.