1. DC Circuits Resistors in Series and Parallel resistor: resistors connected in series: battery: + - V R A B Put your finger on the wire at A. If you can move along the wires to B without ever having a choice of which wire to follow, the circuit components are connected in series. A resistor is any circuit element that has electrical resistance (heater, light bulb, etc.). Usually we assume wires have no resistance.

2. Here’s a circuit with three resistors and a battery: R3R3 R2R2 R1R1 + - V I current flows in the steady state, the same current flows through all resistors III there is a potential difference (voltage drop) across each resistor V1V1 V3V3 V2V2

3. An electric charge q is given a potential energy qV by the battery. R3R3 R2R2 R1R1 + - V I III V1V1 V3V3 V2V2 As it moves through the circuit, the charge loses potential energy qV 1 as it passes through R 1, etc. The charge ends up where it started, so the total energy lost must equal the initial potential energy input: qV = qV 1 + qV 2 + qV 3.

4. R3R3 R2R2 R1R1 + - V I III V1V1 V3V3 V2V2 qV = qV 1 + qV 2 + qV 3 V = V 1 + V 2 + V 3 V = IR 1 + IR 2 + IR 3 Now imagine replacing the three resistors by a single resistor, having a resistance R such that it draws the same current as the three resistors in series.

5. R eq + - V I V I As above:V = IR eq From before:V = IR 1 + IR 2 + IR 3 Combining:IR eq = IR 1 + IR 2 + IR 3 R eq = R 1 + R 2 + R 3 For resistors in series, the total resistance is the sum of the separate resistances.

6. We can generalize this to make an OSE: OSE:R eq =  R i (resistors in series) resistors connected in parallel: AB Put your finger on the wire at A. If in moving along the wires to B you ever having a choice of which wire to follow, the circuit components are connected in parallel. a consequence of conservation of energy

7. V V V R3R3 R2R2 R1R1 + - V I current flows different currents flows through different resistors the voltage drop across each resistor is the same I3I3 I1I1 I2I2 Caution: circuits which are drawn to appear very different may be electrically equivalent.

8. V V V R3R3 R2R2 R1R1 + - V I I3I3 I1I1 I2I2 AB In the steady state, the current I “splits” into I 1, I 2, and I 3 at point A. I I 1, I 2, and I 3 “recombine” to make a current I at point B. Therefore, the net current flowing out of A and into B is I = I 1 + I 2 + I 3. Because the voltage drop across each resistor is V:

9. Now imagine replacing the three resistors by a single resistor, having a resistance R such that it draws the same current as the three resistors in parallel. V R eq + - V I I AB I From above, I = I 1 + I 2 + I 3, and So that

10. Dividing both sides by V gives We can generalize this to make an OSE: OSE:(resistors in parallel) a consequence of conservation of charge

11. Examples How much current flows from the battery in the circuit shown? What is the current through the 500  resistor? + - 12 V 400  500  700  II I1I1 I2I2 a b c

12. What is your stragegy? Step 1—replace the 500 and 700  parallel combination by a single equivalent resistor. + - 12 V 400  500  700  II I1I1 I2I2 a b c Woe is me, what to do? I = ? I 1 = ? Woe is me, what to do? Always think: bite-sized chunks!

13. Step 2—replace the 400 and R eq1 series combination by a single equivalent resistor R eq, net. + - 12 V 400  R eq1 II a b c Woe is me, what to do? I = ? I 1 = ? Woe is me, what to do? Find another bite-sized chunk!

14. + - 12 V R eq1, net II a c Step 3—Solve for the current I. This isn’t so complicated!

15. Step 4—To get I 1, Calculate V bc. + - 12 V 400  500  700  II I1I1 I2I2 a b c Use V total = V ab + V bc. You know V total = V and I so you can get V ab and then V bc. Knowing I, Calculate I 1. Woe is me! Stuck again! V bc V ab

16. + - 12 V 400  500  700  II I1I1 I2I2 a b c The voltage drop across both the 500 and 700  resistors is the same, and equal to V bc. Use V = IR to get I 1 across the 500  resistor.

17. 19.2 EMF and Terminal Voltage We have been making calculations with voltages from batteries without asking detailed questions about the batteries. Now it’s time to ask those questions. We introduce a new term – emf – in this section. Any device which transforms a form of energy into electric energy is called a “source of emf.” “emf” is an abbreviation for “electromotive force,” but emf does not really refer to force! The emf of a source is the voltage it produces when no current is flowing.

18. The voltage you measure across the terminals of a battery (or any source of emf) is less than the emf because of internal resistance. Here’s a battery with an emf. All batteries have an “internal resistance:” + - a b The “battery” is everything inside the green box. Hook up a voltmeter to measure the emf: + - a b The “battery” is everything inside the green box. Getting ready to connect the voltmeter.

19. You can’t measure voltage without some (however small) current flowing, so you can’t measure emf directly. Measuring the emf??? + - a b The “battery” is everything inside the green box. As soon as you connect the voltmeter, current flows. You can only measure V ab. I

20. We model a battery as producing an emf, , and having an internal resistance r: + - a b The “battery” is everything inside the green box. The terminal voltage, V ab, is the voltage you measure with current flowing. When a current I flows through the battery, V ab is related to the emf, , by r 

21. Why the  sign? If the battery is delivering current, the V it delivers is less than the emf, so the – sign is necessary. If the battery is being charged, you have to “force” the current through the battery, and the V to “force” the current through is greater than the emf, so the + sign is necessary. This will become clear as you work (and understand) problems. Operationally, you simply include an extra resistor to represent the battery resistance, and label the battery voltage as an emf instead of V (units are still volts).

22. Example For the circuit below, calculate the current drawn from the battery, the terminal voltage of the battery, and the current in the 6  resistor. 5  4  8  10  6  0.5  = 9 V The following is a “conceptual” solution. Please go back and put in the numbers for yourself.

23. In the next section, we will learn a general technique for solving circuit problems. For now, we break the circuit into manageable bits. “Bite-sized chunks.” Replace the parallel combination by its equivalent. 5  4  8  10  6  0.5  = 9 V Do you see any bite-sized chunks that are simple series or parallel?

24. 5  4  8  10  6  0.5  = 9 V Replace the series combination by its equivalent. Any more “bite-sized chunks?” Pretend that everything inside the green box is a single resistor.

25. 5  4  8  10  6  0.5  = 9 V You are left with an equivalent circuit of 3 resistors in series, which you can handle. Next bite-sized chunk. Inside the blue box is “a” resistor. Replace the parallel combination by its equivalent.