Chapter 20

Electric Circuits

Example. The figure shows a circuit composed of a 24-V battery and four resistors, whose resistances are 110, 180, 220 and 250 Ω. Find (a) the total current supplied by the battery and (b) the voltage between points A and B in the circuit.

Series R = R1 + R2 and Parallel 1/R = 1/R1 + 1/R2

20.8 Circuits Wired Partially in Series and Partially in Parallel

Series: R = 220 Ω + 250 Ω

= 470 Ω

Parallel: 1/R = 1/(180 Ω) + 1/(470 Ω)

R = 130 Ω

Series: R = 110 Ω + 130 Ω

= 240 Ω

Example -- continued

(a) Total current supplied by battery: I = V/R , V = 24 V, R = 240 Ω (equivalent resistance of the circuit) è I = 24/240 = 0.10 A (b) Voltage between points A and B: VAB = IRAB = (0.10)(130) = 13 V

20.9 Internal Resistance

Batteries and generators add some resistance to a circuit. This resistance is called internal resistance. The actual voltage between the terminals of a battery is known as the terminal voltage.

20.9 Internal Resistance

Example 12 The Terminal Voltage of a Battery The car battery has an emf of 12.0 V and an internal resistance of 0.0100 Ω. What is the terminal voltage when the current drawn from the battery is (a) 10.0 A and (b) 100.0 A?

(a) ( )( ) V 10.0 010.0A 0.10 =Ω== IrV

11.9VV 10.0V 0.12 =−

(b) ( )( ) V 0.1 010.0A 0.100 =Ω== IrV

11.0VV 0.1V 0.12 =−

Why do your headlights dim when you start up your car?

r = 0.010 Ω

starter

headlights

A B

Using the numbers in the previous example: Before turning on the starter switch, I1 = 0 A, I = I2 = 10 A and VAB = 11.9 V as in part (a). After turning on the starter switch, a large current flows through I1 so I = 100 A and VAB = 11.0 V as in part (b). Since the headlights see a reduced VAB, they dim. ( Pheadlights = VAB

2/R2 )

12 V

The measurement of current and voltage in DC circuits Devices used: Ammeter -- measures current flowing in the circuit Voltmeter -- measures voltage across some device in the circuit Ammeters and voltmeters can be either analog (read out with the deflection of a needle) or digital devices. We will study how the analog devices work since they’re easier to understand from basic principles. Galvanometer -- an analog device that responds to electrical currents flowing through it by causing a needle to deflect across some scale. Both the ammeter and voltmeter are based on a galvanometer.

20.11 The Measurement of Current and Voltage

How a galvanometer works. The coil of wire and pointer rotate when there is a current in the wire. This one is calibrated so that if a current of 0.10 mA flows through the galvanometer coil, a full-scale deflection of the needle on the calibrated scale Is obtained.

Schematic representation of a galvanometer showing the resistance in its coil RC in series with the galvanometer.

20.11 The Measurement of Current and Voltage

Using the galvanometer as an ammeter. If a galvanometer with a full-scale limit of 0.100 mA is to be used to measure the current of 60.0 mA, a parallel shunt resistance must be used so that the excess current of 59.9 mA can detour around the galvanometer coil. Assuming RC = 50 Ω, find R. VAB = IGRC = (0.1 x 10-3)(50) = 5 x 10-3 V R = VAB/IR = (5 x 10-3)/(59.9 x 10-3) = 8.35 x 10-2 Ω

20.11 The Measurement of Current and Voltage

An ammeter must be inserted into a circuit so that the current passes directly through it. Thus, it is important that it has as low a resistance as possible so as to minimize its effect on the circuit since it acts as a series resistor added to the circuit. Find the equivalent resistance of the ammeter in our example to see if it is small.

RC and R are in parallel, and RC = 50 Ω and R = 8.35 x 10-2 Ω 1/RA = 1/RC + 1/R = 1/(50) + 1/(8.35 x 10-2) = 12.0 Ω-1 --> RA = 0.083 Ω Circuits generally have resistances much higher than this, ~102 to ~103 Ω

Using the galvanometer as a voltmeter

A B

VAB

R

A B

ê

VAB

If a galvanometer with a full-scale limit of 0.100 mA is to be used to measure the voltage of 10.0 V, a series resistor must be used to limit the current flowing through the galvanometer to 0.100 mA. Assuming RC = 50 Ω, find R. RV = R + RC , since R and RC in series VAB = IRV = I(R+RC) R = VAB/I - RC = (10.0)/(0.100 x 10-3) - 50 = 105 Ω

20.11 The Measurement of Current and Voltage

To measure the voltage between two points in a circuit, a voltmeter is connected between the points. Thus, it is important that it has as high a resistance as possible so as to minimize its effect on the circuit since it acts as a parallel resistor added to the circuit. Find the equivalent resistance of the voltmeter in our example to see if it is large.

R and RC are in series, and R = 105 Ω and RC = 50 Ω RV = R + RC = 105 + 50 = 105 Ω Circuits generally have resistances much lower than this, ~102 to ~103 Ω