Chapter 19
Electric Circuits
Kirchhoff’s Rules
How can one deal with a complicated circuit like this?
à Use Kirchhoff’s rules:
* Junction rule
* Loop rule
Kirchhoff’s Rules
The junction rule states that the total
current directed into a junction must
equal the total current directed out of
the junction.
Kirchhoff’s Rules
The loop rule expresses conservation of energy in terms of
the electric potential and states that for a closed circuit loop,
the total of all potential rises is the same as the total of all
potential drops.
Kirchhoff’s Rules
KIRCHHOFF’S RULES
Junction rule. The sum of the magnitudes of the currents directed
into a junction equals the sum of the magnitudes of the currents directed
out of a junction.
Loop rule. Around any closed circuit loop, the sum of the potential drops
equals the sum of the potential rises, i.e. the sum over all of the changes
in potential is zero.
Kirchhoff’s Rules
Example: Using Kirchhoff’s Loop Rule
Determine the current in the circuit.
Kirchhoff’s Rules
Loop around circuit
in clockwise direction.
24 V − I (12 Ω) − 6.0 V − I (8.0 Ω) = 0
I (12 Ω) + 6.0 V + I (8.0 Ω) = 24
V

 potential rises
potential drops
I ( 20 Ω) = 18 V
I = 0.90 A
Kirchhoff’s Rules
Example: A car’s headlight is connected to the battery which is charged
by the alternator as shown in the figure and circuit diagram. Find the
currents through the headlight, battery and alternator.
Kirchhoff’s Rules
Reasoning Strategy in applying Kirchhoff’s Rules
1.  Draw the current in each branch of the circuit. Choose any direction.
If your choice is incorrect, the value obtained for the current will turn out
to be a negative number.
2.  Mark each resistor with a + at one end and a – at the other end in a way
that is consistent with your choice for current direction in step 1. For batteries,
the signs will be the usual + for higher potential and – for lower potential.
3.  Apply the junction rule and the loop rule to the circuit, obtaining in the process
as many independent equations as there are unknown variables.
4. Solve these equations simultaneously for the unknown variables.
rA = 0.100 Ω rB = 0.010 Ω RH = 1.20 Ω
Apply the Junction rule at B:
IA + IB = IH
Eq. 1
Apply the Loop rule for ABEF and BCDE
In the clockwise direction:
ABEF:
BCDE:
2 + I B rB
14 − I A rA + I B rB −12 = 0 ⇒ I A =
rA
12 − I B rB
12 − I B rB − I H RH = 0 ⇒ I H =
RH
Eq. 2
Eq. 3
Put Eq. 2 and Eq. 3 into Eq. 1, solve for IB. Then use Eq. 2 and Eq. 3 to get IA and IH:
2 + I B rB
12 − I B rB
+ IB =
rA
RH
⇒ IB =
12rA − 2RH
= −9.02 A I A = 19.10 A I H = 10.08 A
rB RH + rA RH + rA rB
IB flows in opposite direction of diagram
Capacitors in Series and Parallel
q = q1 + q2 = C1V + C2V = (C1 + C2 )V
Parallel capacitors
CP = C1 + C2 + C3 + 
Capacitors in Series and Parallel
⎛ 1
q
q
1 ⎞
V = V1 + V2 = +
= q⎜⎜ + ⎟⎟
C1 C2
⎝ C1 C2 ⎠
Series capacitors
1
1
1
1
= +
+
+
CS C1 C2 C3
Example: Two capacitors have values 2.7 µF and 4.2 µF. Find their
equivalent capacitance if they are connected in a) series, and b) parallel
with each other. If the capacitive networks in a) and b) are connected
across a 200 V source, find c) the energy stored in them in each case.
a)  Series
1
1
1
5 −1
=
+
=
6.1×10
F
−6
−6
Cs 2.7 ×10
4.2 ×10
b)  Parallel C p = 2.7 ×10 −6 + 4.2 ×10 −6 = 6.9 ×10 −6 F
1
CV 2
2
1
2
−6
Energys = (1.6 ×10 ) ( 200 ) = 0.032 J
2
1
2
−6
Energy p = ( 6.9 ×10 ) ( 200 ) = 0.14 J
2
c)  Energy =
Cs = 1.6 ×10 −6 F
RC Circuits
Capacitor charging
−t RC $
"
q = qo #1− e
%
time constant
τ = RC
RC Circuits
Capacitor discharging
q = qo e −t RC
time constant
τ = RC
Example: How long does it take to discharge an RC circuit
to the fraction 1/e (about 37%) of its original charge for a
1000 Ω resistor connected in series with a) a 500 µF capacitor,
and b) a 0.5 F capacitor?
a) τ = RC = (1000 ) 50 ×10 −6 = 0.05 s = 50 ms
(
)
b) τ = RC = (1000 ) ( 0.5) = 500 s = 8.3 min
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.
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.
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
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
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
Safety in grounding electrical devices
To reduce the danger inherent in using circuits, proper electrical grounding
is necessary.