AP Physics Section 19-1 to 19-3 Simple DC Circuits

EMF and terminal voltage Electrochemical cells and batteries are not constant current sources, since current varies with the resistance load. However, they are nearly constant sources of potential difference. Cells do have an internal resistance that becomes evident when current flows through them. This internal resistance, r, arises because the chemical reactions that create the EMF (voltage) can only occur at a finite speed. Charged particles that take part in the reactions have to physically move between electrodes. Also, as the cell ages, products build up that slow reactions.

When testing a cell with a voltmeter, the current flow is almost zero, and the voltage reading is close to the true EMF. However, current flow through the cell reduces the potential different by the amount ∆Vint = Ir. alkaline cell Terminal voltage of a cell with a load is given by: – + a b I ∆Vab = ℰ – Ir ℰ r

+ – – + ℰ = lead-acid battery a b 0.5 Ω 12.0 V A 65.0 Ω resistor is connected to the terminals of a lead-acid auto battery. The EMF of the battery is 12.0 V and the internal resistance is 0.5 Ω. Calculate the current, the terminal voltage, and power dissipated in the two resistors.

ℰ ∆Vab = ℰ – Ir ∆Vab = IR and Ohm’s Law IR = ℰ – Ir ℰ = give: so, Finding Current: ∆Vab = ℰ – Ir ∆Vab = IR and Ohm’s Law IR = ℰ – Ir ℰ = give: so, IR + Ir ℰ = I(R + r) ℰ R + r 12.0 V Therefore: I = = 65.0 Ω + 0.5 Ω I = 0.183 A Terminal voltage: ∆Vab = ℰ – Ir ∆Vab = 12.0 V – (0.183 A)(0.5 Ω) ∆Vab = 11.9 V Power dissipated: In R: PR = I2R = (0.183 A)2(65.0 Ω) = 2.18 W In r: Pr = I2r = (0.183 A)2(0.5 Ω) = 0.02 W

Resistors in series ∆VS = ∆V1 + ∆V2 + ∆V3 ∆VS = IR1 + IR2 + IR3 R1 R2 We saw in the last example that the total resistance in the circuit equalled the sum of R and r. This is because the voltage drops by a certain amount across each resistor. The sum of the drops equals the total potential difference: ∆VS = ∆V1 + ∆V2 + ∆V3 The current, I, is the same through all resistors so: ∆VS = IR1 + IR2 + IR3 I R1 R2 R3 ∆VS V1 V2 V3

∑ ∆VS = IRS Since ∆VS = I(R1 + R2 + R3) RS = R1 + R2 + R3 RS = Ri i n According to Ohm’s Law, the voltage for the whole series circuit should be: ∆VS = IRS where RS is the total equivalent resistance in the circuit. Since ∆VS = I(R1 + R2 + R3) we can see that the total equivalent resistance is: RS = R1 + R2 + R3 RS = ∑ Ri i n In general: on AP sheet

Resistors in parallel I1 ∆VS I2 I3 When resistors are placed in parallel, the potential across each reistor in the same. The ends of each resistor are connected. However, the current flow divides between the parallel branches, reducing the current in each branch. The sum of the currents through the branches has to equal the total current flowing into them: I I = I1 + I2 + I3

∑ ∆V ∆V ∆V I1 = I2 = I3 = R1 R2 R3 I = ∆V RP ∆V RP = R1 + R2 R3 1 RP = According to Ohm’s Law, the current in each branch should be: ∆V ∆V ∆V I1 = I2 = I3 = R1 R2 R3 Also according to Ohm’s Law, the current in the parallel resistor circuit as a whole is: I = ∆V RP Substituting all of these into the total current equation yields: ∆V RP = R1 + R2 R3 Dividing by ∆V gives: on AP sheet In general: 1 RP = R1 + R2 R3 = ∑ i n 1 RP Ri

Combined resistor circuits Calculate the total circuit resistance, the total current, and the voltage and current across every resistor. 9V 1 R45 = R4 + R5 1 37 + 45 = = 0.0270 + 0.0222 1 R45 = 0.0492 R45 = 20.3 Ω R456 = R45 + R6 = 20.3 + 75 = 95.3 Ω R23 = R2 + R3 = 49 + 51 = 100 Ω

1 R23456 = R23 + R456 1 100 + 95.3 = = 0.0100 + 0.0105 1 R23456 = 0.0205 R23456 = 48.8 Ω RT = R1 + R23456 = 80 + 48.8 = 129 Ω I = ∆V RT 9V = = 0.0698 A I1 = 0.0698 A 129 Ω V1 = I1R1 = (0.0698 A)(80 Ω) = 5.58 V drop

V23456 = IR23456 = (0.0698 A)(48.8 Ω) = 3.41 V drop ∆V23 I2 = 3.41 V It should drop the rest of the 9 V across the remaining resistors, but we can prove that: V23456 = IR23456 = (0.0698 A)(48.8 Ω) = 3.41 V drop Check: 5.58 V + 3.41 V = 8.99 V close enough! Now find the currents in the upper and lower branches: ∆V23 I2 = 3.41 V 0.0341 A I23 = = = 0.0341 A I3 = R23 100 Ω 0.0341 A The current in the lower branch should be the remaining portion of total I, but we can prove it: ∆V456 3.41 V I456 = = = 0.0358 A R456 95.3 Ω I6 = 0.0358 A Check: 0.0341 A + 0.0358 A = 0.0699 A close enough!

I23 = I2 = I3 = V2 = I2R2 = (0.0341 A)(49 Ω) = 1.67 V drop V3 = I3R3 = Voltage drops across R2 and R3: I23 = I2 = I3 = 0.0341 A V2 = I2R2 = (0.0341 A)(49 Ω) = 1.67 V drop V3 = I3R3 = (0.0341 A)(51 Ω) = 1.74 V drop 3.41 V These should add up to 3.41 V: 1.67 V + 1.74 V = Voltage drops across R45 and R6: I456 = I45 = I6 = 0.0358 A V45 = I45R45 = (0.0358 A)(20.3 Ω) = 0.727 V drop V6 = I6R6 = (0.0358 A)(75 Ω) = 2.69 V drop These should also add up to 3.41 V: 0.727 V + 2.69 V = 3.42 V close enough!

V4 = V5 = 0.727 V drop ∆V4 0.727 V I4 = = = R4 37 Ω ∆V5 0.727 V I5 = = Voltage drops across R4 and R5 are the same: V4 = V5 = 0.727 V drop Lastly, find the current through R4 and R5: ∆V4 0.727 V I4 = = = 0.0196 A R4 37 Ω ∆V5 0.727 V I5 = = = 0.0162 A R5 45 Ω These she add up to the current in the lower branch, 0.0358 A: 0.0196 A + 0.0162 A = 0.358 A We now know the current through every resistor and the potential difference across each! Yay!

Rule 1: The junction rule From these types of simple circuits G. R. Kirchhoff devised two rules to handle more complex circuits. Rule 1: The junction rule At any junction point, the sum of all the currents entering the junction must equal the sum of all currents leaving the junction. This is based on the conservation of electric charge. Rule 2: The loop rule The sum of the changes in potential around ANY closed path must be zero. This is based on the conservation of energy.