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Published byGalilea Toye Modified over 2 years ago

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a b R C I I R R I I r V

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Resistors in ParallelParallel a d I I R1R1 R2R2 I1I1 I2I2 V I a d I RV But current through R 1 is not I ! Call it I 1. Similarly, R 2 I 2. KVL What to do? Very generally, devices in parallel have the same voltage drop How is I related to I 1 & I 2 ? Current is conserved!

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Another (intuitive) way… Consider two cylindrical resistors with cross-sectional areas A 1 and A 2 V R1R1 R2R2 A1A1 A2A2 Put them together, side by side … to make one “fatter”one,

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Consider the circuit shown: –What is the relation between V a - V d and V a - V c ? (a) (V a -V d ) < (V a -V c ) (b) (V a -V d ) = (V a -V c ) (c) (V a -V d ) > (V a -V c ) 12V I1I1 I2I2 a b d c 50 20 80 (a) I 1 < I 2 (b) I 1 = I 2 (c) I 1 > I 2 1B – What is the relation between I 1 and I 2 ? 1B 1A

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Consider the circuit shown: –What is the relation between V a - V d and V a - V c ? (a) (V a -V d ) < (V a -V c ) (b) (V a -V d ) = (V a -V c ) (c) (V a -V d ) > (V a -V c ) 12V I1I1 I2I2 a b d c 50 20 80 1A Do you remember that thing about potential being independent of path? Well, that’s what’s going on here !!! ( V a - V d ) = ( V a - V c ) Point d and c are the same, electrically

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(a) I 1 < I 2 (b) I 1 = I 2 (c) I 1 > I 2 1B – What is the relation between I 1 and I 2 ? 1B Consider the circuit shown: –What is the relation between V a - V d and V a - V c ? (a) (V a -V d ) < (V a -V c ) (b) (V a -V d ) = (V a -V c ) (c) (V a -V d ) > (V a -V c ) 12V I1I1 I2I2 a b d c 50 20 80 1A Note that: V b -V d = V b -V c Therefore,

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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, I i assigned to it

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

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How to use Kirchhoff’s Laws A two loop example: Analyze the circuit and identify all circuit nodes and use KCL. (2) 1 I 1 R 1 I 2 R 2 = 0 (3) 1 I 1 R 1 2 I 3 R 3 = 0 (4) I 2 R 2 2 I 3 R 3 = 0 1 2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3 (1) I 1 = I 2 + I 3 Identify all independent loops and use KVL.

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How to use Kirchoff’s Laws 1 2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3 Solve the equations for I 1, I 2, and I 3 : First find I 2 and I 3 in terms of I 1 : From eqn. (2) From eqn. (3) Now solve for I 1 using eqn. (1):

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Let’s plug in some numbers 1 2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3 1 = 24 V 2 = 12 V R 1 = 5 R 2 =3 R 3 =4 Then I 1 =2.809 A and I 2 = A and I 3 = A See Appendix for a more complicated example, with three loops. The sign means that the direction of I 3 is opposite to what’s shown in the circuit

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

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

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

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Power Batteries & Resistors Energy expended 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

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Power Transmission Why do we use “high tension” lines to transport power? – Note: for fixed input power and line resistance, the inefficiency 1/V 2 Example: Quebec to Montreal 1000 km R= 220 suppose P in = 500 MW With V in =735kV, = 80%. The efficiency goes to zero quickly if V in were lowered! Keep R small Make V in big –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: 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

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Resistor- capacitor circuits R C I I 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…

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Capacitors Circuits, Qualitative Charging (it takes time to put the final charge on) –Initially, the capacitor behaves like a wire ( V = 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. Basic principle: Capacitor resists rapid change in Q resists rapid changes in V

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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 I 0+ just after the switch is thrown? (a) I 0+ = 0 (b) I 0+ = /2 R (c) I 0+ = 2 / R a b R C I I R 2A (a) I = 0 (b) I = /2 R (c) I > 2 / R – What is the value of the current I after a very long time? 2B

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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 I 0+ just after the switch is thrown? (a) I 0+ = 0 (b) I 0+ = /2 R (c) I 0+ = 2 / R a b R C I I R 2A 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+, IR 0 IR = 0 I = /2 R

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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 I 0+ just after the switch is thrown? 2A (a) I 0+ = 0 (b) I 0+ = /2 R (c) I 0+ = 2 / R (a) I = 0 (b) I = /2 R (c) I > 2 / 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 ; the current goes to 0. a b R C I I R

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Summary Kirchhoff’s Laws –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

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–determine which KCL equations are algebraically independent (not all are in this circuit!) –I 1 = I 2 + I 3 –I 4 = I 2 + I 3 –I 4 = I 5 + I 6 –I 1 = I 5 + I 6 Appendix: A three-loop KVL example Identify all circuit nodes - these are where KCL eqn’s are found Analyze circuit and identify all independent loops where V i = 0 KVL I1=I4I1=I2+I3I4+I5+I6I1=I4I1=I2+I3I4+I5+I6 I1I1 I2I2 I3I3 I4I4 I5I5 I6I6

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A three-loop KVL example Now, for Kirchoff’s voltage law: (first, name the resistors) I1=I4I1=I2+I3I4+I5+I6I1=I4I1=I2+I3I4+I5+I6 I1I1 I2I2 I3I3 I4I4 I5I5 I6I6 Here are the node equations from applying Kirchoff’s current law: There are simpler ways of analyzing this circuit, but this illustrates Kirchoff’s laws R1R1 R 2a R 2b R3R3 R4R4 R6R6 R5R5 I 6 R 6 +I 5 R 5 =0 I 2 R 2b +I 2 R 2a - I 3 R 3 =0 V B -I 1 R 1 -I 2 R 2a - I 2 R 2b -I 4 R 4 -I 5 R 5 = 0 Six equations, six unknowns….

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