Lecture 12: TUE 02 FEB DC circuits Ch27.4-9 Physics 2102 Jonathan Dowling.

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Lecture 12: TUE 02 FEB DC circuits Ch Physics 2102 Jonathan Dowling

How to Solve Multi-Loop Circuits

Step I: Simplify “Compile” Circuits Resistors Capacitors Key formula: V=iR Q=CV In series: same current dQ/dt same charge Q R eq =∑R j 1/C eq = ∑1/C j In parallel: same voltage same voltage 1/R eq = ∑1/R j C eq =∑C j P = iV = i 2 R = V 2 /R U = QV/2 = Q 2 /2C = CV 2

Note: Skip Compile and Go Straight to Loop and Junction Rules if Number of Batteries is Large One Battery: Compile First Three Batteries: Straight to Loop & Junction

Around every loop add + E if you cross a battery from minus to plus, – E if plus to minus, and –iR for each resistor. Then sum to Zero: + E 1 – E 2 – iR 1 – iR 2 = 0. Step II: Apply Loop Rule R1R1 R2R2 E1E1 E2E2 +–+– +–+– Conservation of ENERGY!

Step II: Apply Junction Rule At every junction sum the ingoing currents and outgoing currents and set them equal. i 1 = i 2 + i 3 i1i1 i2i2 i3i3 Conservation of CHARGE!

Step III: Equations to Unknowns Continue Steps I–III until you have as many equations as unknowns! + E 1 – E 2 – i 1 R 1 – i 2 R 2 = 0 and i = i 1 + i 2 Solve for i 2, i 3 Given: E 1, E 2, i, R 1, R 2

Example Find the equivalent resistance between points (a) F and H and (b) F and G. (Hint: For each pair of points, imagine that a battery is connected across the pair.) Compile R’s in Series Compile equivalent R’s in Parallel F H F H F H Slide Rule Series Parallel

Example Assume the batteries are ideal, and have emf E 1 =8V, E 2 =5V, E3=4V, and R 1 =140  R 2 =75  and  R 3 =2 . What is the current in each branch? What is the power delivered by each battery? Which point is at a higher potential, a or b? Apply loop rule three times and junction rule twice.

Example What’s the current through resistor R 1 ? What’s the current through resistor R 2 ? What’s the current through each battery? Apply loop rule three times and junction rule twice.

Non-Ideal Batteries You have two ideal identical batteries, and a resistor. Do you connect the batteries in series or in parallel to get maximum current through R? Does the answer change if you have non-ideal (but still identical) batteries? Apply loop and junction rules until you have current in R.

More Light Bulbs If all batteries are ideal, and all batteries and light bulbs are identical, in which arrangements will the light bulbs as bright as the one in circuit X? Does the answer change if batteries are not ideal? Calculate i and V across each bulb. P = iV = “brightness” or Calculate each i with R’s the same: P = i 2 R

i(t) E /R RC Circuits: Charging a Capacitor In these circuits, current will change for a while, and then stay constant. We want to solve for current as a function of time i(t)=dq/dt. The charge on the capacitor will also be a function of time: q(t). The voltage across the resistor and the capacitor also change with time. To charge the capacitor, close the switch on a. Time constant:  =RC A differential equation for q(t)! The solution is: CECE V C =Q/C V R =iR

i(t) E /R RC Circuits: Discharging a Capacitor Assume the switch has been closed on a for a long time: the capacitor will be charged with Q=CE. Then, close the switch on b: charges find their way across the circuit, establishing a current. C +  +++  Solution:

Example The three circuits below are connected to the same ideal battery with emf E. All resistors have resistance R, and all capacitors have capacitance C. Which capacitor takes the longest in getting charged? Which capacitor ends up with the largest charge? What’s the final current delivered by each battery? What happens when we disconnect the battery? Compile R’s into into R eq. Then apply charging formula with R eq C = 

Example In the figure, E = 1 kV, C = 10 µF, R 1 = R 2 = R 3 = 1 M . With C completely uncharged, switch S is suddenly closed (at t = 0). What’s the current through each resistor at t=0? What’s the current through each resistor after a long time? How long is a long time? Compile R 1, R 2, and R 3 into R eq. Then apply discharging formula with R eq C = 