Chapter 31 – Resistor Circuits

This chapter synthesizes material from earlier chapters. What happens when resistors, batteries, and capacitors are linked together to form a circuit? Circuits are analyzed by using Kirchoff’s two laws: the total current entering a junction equals the total current leaving a junction and the sum of potential drops going around a loop are zero. Of course as you ought to have already learned, stating a law is not the same as using it to actually solve a problem!

1. For example, when traversing a loop in the direction of the current (guessed in many cases), the potential drop across a resistor is always negative, -IR. Charges lose energy passing through a resistor. On the other hand, the change in emf going through a battery is positive if while traveling around the loop the imaginary charge passes through the battery from negative to positive (increase in potential) or negative if the charge moves from the positive to negative terminal (decrease in potential). Using Kirchoff’s laws correctly involves practice and skill in solving simultaneous equations. None of the problems in this chapter demand a high level of skill in using Kirchoff’s laws.

2. The power output of a battery is I E, current times emf. The power expended going through a resistor is I²R and shows up as heat or a combination of heat and light for an incandescent light bulb. The heat is generated because of all the collisions between the drifting electrons and the metal ions in the wire/resistor. Capacitors do not dissipate or generate power but instead store energy.

3. For two resistors hooked in series with a battery, ΔV_bat, by Kirchoff’s loop law, the net potential change through the two resistors is just the sum IR₁ + IR₂ = ΔV_bat. That potential drop is equal to the potential drop through the equivalent resistor used to replace the two resistors. And since the current through the equivalent resistor is the same, IR _eq = IR₁ + IR₂ or R _eq = R₁ + R₂.

4. For resistors in parallel, the potential drop across each of the individual resistors is the same as the potential drop across the equivalent resistor and the current through the equivalent resistor is the sum of the current through the individual resistors, I_eq = I₁ + I₂ or ΔV/R_eq = ΔV/R₁ + ΔV/R₂. Dividing by ΔV gives,

1/R_eq = 1/R₁ + 1/R₂, for resistors in parallel, the equivalent resistance is found by adding reciprocals.

5. Ammeters measure current passing through a part of a circuit. To this end, the current being measured has to pass through the ammeter. Since the ammeter has to have some internal resistance, the current passing through the ammeter along the part of the circuit of interest must now be less than it was before! The bigger the internal resistance of the ammeter, the bigger effect the ammeter will have on the circuit.

6. Voltmeters measure the voltage change across an element of a circuit. Since the voltmeter is measuring voltage difference it has to be connected in parallel with the element of interest. Now the current heading to the element has two options, it can pass through the element, good, or can pass through the voltmeter, bad. Since the potential drop measured by the voltmeter depends on the current going through the element of interest, the voltmeter will measure a voltage difference less than the voltage drop that existed before the voltmeter was inserted in the circuit.

7. Real batteries have some internal resistence. Consequently the potential available from a real battery is E - I r. The more current, the larger the difference between the batteries listed emf and the actual voltage difference across the terminals.

8. For a capacitor and resistor hooked in series, the time it takes for a charged capacitor to discharge or an uncharged capacitor to charge both depend on the product of R and C, circuit’s time constant = RC.

Discharging: Q(t) = Q_max e^-t/RC

Charging: Q(t) = Q_max(1 - e^-t/RC)