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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 = Ir = (10.0 A )(0.010 Ω ) = 0.10 V
12.0 V − 0.10 V = 11.9V
(b)
V = Ir = (100.0 A )(0.010 Ω ) = 1.0 V
12.0 V − 1.0 V = 11.0V
Why do your headlights dim when you start up your car?
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 = VAB2/R2 )
headlights
A
B
starter
r = 0.010 Ω
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
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.
A
VAB
RV = R + RC , since R and RC in series
B
VAB = IRV = I(R+RC)
ê
R = VAB/I - RC = (10.0)/(0.100 x 10-3) - 50
= 105 Ω
R
A
VAB
B
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