# Kirchhoff’s Laws a b e R C I e R I r V.

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Kirchhoff’s Laws a b e R C I e R I r V

Resistors in Parallel I R1 R2 I1 I2 V I R V a d KVL Þ a d Þ
What to do? Very generally, devices in parallel have the same voltage drop a d I R1 R2 I1 I2 V But current through R1 is not I ! Call it I1. Similarly, R2 «I2. KVL Þ I a d R V How is I related to I 1 & I 2 ? Current is conserved! Þ

Another (intuitive) way…
Consider two cylindrical resistors with cross-sectional areas A1 and A2 V R1 R2 A1 A2 Put them together, side by side … to make one “fatter”one, Þ

Consider the circuit shown:
What is the relation between Va -Vd and Va -Vc ? 12V I1 I2 a b d c 50W 20W 80W 1A (a) (Va -Vd) < (Va -Vc) (b) (Va -Vd) = (Va -Vc) (c) (Va -Vd) > (Va -Vc) (a) I1 < I2 (b) I1 = I2 (c) I1 > I2 1B What is the relation between I1 and I2?

Consider the circuit shown:
What is the relation between Va -Vd and Va -Vc ? 12V I1 I2 a b d c 50W 20W 80W 1A (a) (Va -Vd) < (Va -Vc) (b) (Va -Vd) = (Va -Vc) (c) (Va -Vd) > (Va -Vc) Do you remember that thing about potential being independent of path? Well, that’s what’s going on here !!! (Va -Vd) = (Va -Vc) Point d and c are the same, electrically

Consider the circuit shown:
What is the relation between Va -Vd and Va -Vc ? 12V I1 I2 a b d c 50W 20W 80W 1A (a) (Va -Vd) < (Va -Vc) (b) (Va -Vd) = (Va -Vc) (c) (Va -Vd) > (Va -Vc) 1B 1B What is the relation between I1 and I2? (a) I1 < I2 (b) I1 = I2 (c) I1 > I2 Note that: Vb -Vd = Vb -Vc Therefore,

Kirchhoff’s Second Rule “Junction Rule” or “Kirchhoff’s Current Law (KCL)”
In deriving the formula for the equivalent resistance of 2 resistors in parallel, we applied Kirchhoff's Second Rule (the junction rule). "At any junction point in a circuit where the current can divide (also called a node), the sum of the currents into the node must equal the sum of the currents out of the node." This is just a statement of the conservation of charge at any given node. The currents entering and leaving circuit nodes are known as “branch currents”. Each distinct branch must have a current, Ii assigned to it

Old Preflight Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open. 2) If switch S is closed, what happens to the brightness of the bulb R2? a) It increases b) It decreases c) It doesn’t change 3) What happens to the current I, after the switch is closed ? a) Iafter = 1/2 Ibefore b) Iafter = Ibefore c) Iafter = 2 Ibefore

How to use Kirchhoff’s Laws
A two loop example: e 1 e 2 R1 R3 R2 I3 I1 I2 Analyze the circuit and identify all circuit nodes and use KCL. (1) I1 = I2 + I3 Identify all independent loops and use KVL. (2) e1 - I1R1 - I2R2 = 0 (3) e1 - I1R1 - e2 - I3R3 = 0 (4) I2R2 - e2 - I3R3 = 0

How to use Kirchoff’s Laws
Solve the equations for I1, I2, and I3: First find I2 and I3 in terms of I1 : From eqn. (2) From eqn. (3) Now solve for I1 using eqn. (1): Þ

Let’s plug in some numbers
e1 = 24 V e 2 = 12 V R1= 5W R2=3W R3=4W Then I1=2.809 A and I2= A and I3= A The sign means that the direction of I3 is opposite to what’s shown in the circuit See Appendix for a more complicated example, with three loops.

Summary of Simple Circuits
Resistors in series: Current thru is same; Voltage drop across is IRi Resistors in parallel: Voltage drop across is same; Current thru is V/Ri Kirchhoff’s laws: (for further discussion see online “tutorial essay”)

Batteries (“Nonideal” = cannot output arbitrary current)
V Parameterized with "internal resistance" Þ

Internal Resistance Demo As # bulbs increases, what happens to “R”??
V R e How big is “r”?

Power Batteries & Resistors Energy expended What’s happening? Assert:
chemical to electrical to heat What’s happening? Assert: Rate is: Charges per time Potential difference per charge Units okay? For Resistors: or you can write it as

Power Transmission Why do we use “high tension” lines to transport power? Transmission of power is typically at very high voltages (e.g., ~500 kV) But why? Calculate ohmic losses in the transmission lines Define efficiency of transmission: Keep R small Make Vin big Note: for fixed input power and line resistance, the inefficiency µ 1/V2 Example: Quebec to Montreal 1000 km Þ R= 220W suppose Pin = 500 MW With Vin=735kV, e = 80%. The efficiency goes to zero quickly if Vin were lowered! Why do we use AC (60 Hz)? Easy to generate high voltage (water/steam → turbine in magnetic field → induced EMF) [Lecture 16] We can use transformers [Lecture 18] to raise the voltage for transmission and lower the voltage for use

Resistor- capacitor circuits
Let’s try to add a Capacitor to our simple circuit Recall voltage “drop” on C? Write KVL: What’s wrong here? Consider that and substitute. Now eqn. has only “Q”: KVL gives Differential Equation ! We will solve this next time. For now, look at qualitative behavior…

Capacitors Circuits, Qualitative
Basic principle: Capacitor resists rapid change in Q  resists rapid changes in V Charging (it takes time to put the final charge on) Initially, the capacitor behaves like a wire (DV = 0, since Q = 0). As current starts to flow, charge builds up on the capacitor  it then becomes more difficult to add more charge  the current slows down After a long time, the capacitor behaves like an open switch. Discharging Initially, the capacitor behaves like a battery. After a long time, the capacitor behaves like a wire.

At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. What is the value of the current I0+ just after the switch is thrown? 2A a I I R b C e R (a) I0+ = 0 (b) I0+ = e /2R (c) I0+ = 2e /R (a) I¥ = 0 (b) I¥ = e /2R (c) I¥ > 2e /R What is the value of the current I¥ after a very long time? 2B

At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. What is the value of the current I0+ just after the switch is thrown? 2A a b e R C I (a) I0+ = 0 (b) I0+ = e /2R (c) I0+ = 2e /R Just after the switch is thrown, the capacitor still has no charge, therefore the voltage drop across the capacitor = 0! Applying KVL to the loop at t=0+, e -IR IR = 0 Þ I = e /2R

2A (a) I0+ = 0 (b) I0+ = e /2R (c) I0+ = 2e /R (a) I¥ = 0
At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. What is the value of the current I0+ just after the switch is thrown? 2A (a) I0+ = 0 (b) I0+ = e /2R (c) I0+ = 2e /R (a) I¥ = 0 (b) I¥ = e /2R (c) I¥ > 2e /R What is the value of the current I¥ after a very long time? 2B The key here is to realize that as the current continues to flow, the charge on the capacitor continues to grow. As the charge on the capacitor continues to grow, the voltage across the capacitor will increase. The voltage across the capacitor is limited to e ; the current goes to 0.

Summary Kirchhoff’s Laws Non-ideal Batteries & Power
KCL: Junction Rule (Charge is conserved) Review KVL (V is independent of path) Non-ideal Batteries & Power Effective “internal resistance” limits current Power generated ( ) = Power dissipated ( ) Power transmission most efficient at low current  high voltage Resistor-Capacitor Circuits Capacitors resist rapid changes in Q  resist changes in V

Appendix: A three-loop KVL example
Identify all circuit nodes - these are where KCL eqn’s are found determine which KCL equations are algebraically independent (not all are in this circuit!) I1=I2+I3 I4=I2+I3 I4=I5+I6 I1=I5+I6 I1=I4 I1=I2+I3 I4+I5+I6 Analyze circuit and identify all independent loops where S DVi = 0 <- KVL

A three-loop KVL example
I1 I2 I3 I4 I5 I6 Here are the node equations from applying Kirchoff’s current law: R1 R2a R2b R3 R4 R6 R5 I1=I4 I1=I2+I3 I4+I5+I6 Now, for Kirchoff’s voltage law: (first, name the resistors) -I6R6+I5R5=0 I2R2b+I2R2a- I3R3 =0 VB-I1R1-I2R2a- I2R2b-I4R4-I5R5 = 0 Six equations, six unknowns…. There are simpler ways of analyzing this circuit, but this illustrates Kirchoff’s laws

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