Chapter 25: Electric Circuits
Resistors in Series and Parallel
‰ Resistors
in series
V
V
IR1 + IR2 = V = IReq ⇒ Req = R1 + R2
In general you can extend this formula to : Req = ∑i Ri
Resistors in Series and Parallel
‰ Resistors
in parallel
V
V
1
1
1
V
V V
= +
⇒
= +
I = I1 + I 2 ⇒
Req R1 R2
Req R1 R2
I1 R2
V = I1 R1 = I 2 R2 ⇒ =
I 2 R1
1
1
= ∑i
In general you can extend this formula to :
Req
Ri
Resistors in Series and Parallel
‰ Example
1:
Resistors in Series and Parallel
‰ Example:
(cont’d)
I2
R2
I4
R4
I3
I
R3
∆V
I = ∆V / Req = 12 V/2 Ω = 6 A
I 3 = ∆V / R3 = 12 V/3 Ω = 4 A
I 2 = I 4 = ∆V /( R2 + R3 ) = 12 V/(2 + 4 Ω) = 2 A
Resistors in Series and Parallel
‰ Example:
(cont’d)
Kirchhoff’s Rules
‰ Introduction
• Many practical resistor networks cannot be reduced to simple series-parallel
combinations (see an example below).
• Terminology:
-A junction in a circuit is a point where three or more conductors meet.
-A loop is any closed conducting path.
junction
Loop 2
i
i
i2
i1
i
Loop 1
i2
i
junction
Kirchhoff’s Rules
‰ Kirchhoff’s
junction rule
• The algebraic sum of the currents into any junction is zero:
∑ I = 0 at any junction
Kirchhoff’s Rules
‰ Kirchhoff’s
loop rule
• The algebraic sum of the potential differences in any loop, including
those associated with emfs and those of resistive elements, must equal
zero.
∑V = 0 for any loop
Kirchhoff’s Rules
‰
Rules for Kirchhoff’s loop rule
∑ I = 0 at any junction
∑V = 0 for any loop
Kirchhoff’s Rules
‰
Rules for Kirchhoff’s loop rule (cont’d)
Kirchhoff’s Rules
‰
Solving problems using Kirchhoff’s rules
Kirchhoff’s Rules
‰
Example 1
Kirchhoff’s Rules
‰
Example 1 (cont’d)
Kirchhoff’s Rules
‰
Example 1 (cont’d)
Kirchhoff’s Rules
Find
all the currents
including directions.
‰ Example
2
Loop 2
i
i
i2
i1
i
Loop 1
i2
i
Loop 1
0 = +8V + 4V − 4V − 3i − 2i1
0 = 8 − 3i1 − 3i 2 − 2i1
0 = 8 − 5i1 − 3i 2
multiply by 2
i = i1+ i2
Loop 2
− 6i 2 + 4 + 2i1 = 0
− 6i 2 + 16 − 10i1 = 0
0 − 12 + 12i1 = 0
i1 = 1A
− 6i2 + 4 + 2(1A) = 0
i 2 = 1A
i = 2A
Electrical Measuring Instruments
‰
Galvanometer
To be discussed in a later
class.
Electrical Measuring Instruments
‰
Ammeter
Electrical Measuring Instruments
‰
Ammeter (cont’d)
Electrical Measuring Instruments
‰
Voltmeter
R-C Circuits
‰
Charging a capacitor
R-C Circuits
‰
Charging a capacitor (cont’d)
R-C Circuits
‰
Charging a capacitor (cont’d)
R-C Circuits
‰
Charging a capacitor (cont’d)
R-C Circuits
‰
Charging a capacitor (cont’d)
R-C Circuits
‰
Discharging a capacitor
R-C Circuits
‰
Discharging a capacitor (cont’d)
R-C Circuits
‰
Discharging a capacitor (cont’d)
Exercises
‰
Problem 1
The resistance of a galvanometer coil is 25.0 Ω,
and the current required for full-scale deflection
is 500 µA.
a) Show in a diagram how to convert the galvanometer to an ammeter reading 20.0 mA full scale,
and compute the shunt resistance.
b) Show how to convert the galvanometer to a
voltmeter reading 500 mV full scale, and compute
the series resistance.
Solution
a) For a 20-mA ammeter, the two resistance are in
parallel:
Vc=Vs->IcRc=IsRs->(500 x 10-6 A)(25.0 Ω) =
(20 x 10-3 A – 500 x 10-6 A)Rs-> Rs=0.641 Ω.
b) For a 500-mV voltmeter, the resistances are in
series:
Vab=I(Rc+Rs)->Rs=Vab/I – Rc ->
Rs=500 x 10-3 V / 500 x 10-6 A – 25.0 Ω = 975 Ω.
Rc=25.0 Ω
500 µA
20 mA
Rs
a) ammeter
Rc=25.0 Ω
Rs
a
Vab=500 mV b
b) voltmeter
500 µA
Exercises
36 − 5I 2 − 4( I 2 − I1 ) = 0 → 36 + 4 I1 − 9 I 2 = 0
Solving these two equations for the currents :
I1 = 5.21 A, I 2 = 6.32 A. The current that goes through
4 Ω is I 2 − I1 = 1.11 A.
5.00 Ω
I2
4.00 Ω
loop 2 (right) :
+
I1
2.00 Ω
loop 1 (left) :
20 − 14 − 2 I1 + 4( I 2 − I1 ) = 0 → 6 − 6 I1 + 4 I 2 = 0
20.0 V
+
v
v
+
I1-I2
14.0 V
v
Problem 2
v
‰
36.0 V
Exercises
6.00 Ω
V=18.0 V
6.00 µF
b
a
S
3.00 Ω
‰ Problem 3
a) What is the potential of point a with respect
to point b when the switch S is open?
b) Which point, a or b, is at higher potential?
Now the switch S is closed.
b) What is the final potential of point b?
c) How much charge flows through switch S
when it is closed?
Solution
a) With an open switch:
Q = CeqV = (2.00 × 10 −6 F)(18.0 V) = 3.60 × 10-5 C.
Also, there is a current in the left branch:
I = (18.0 V)/(6.00 Ω + 3.00 Ω) = 2.00 A.
So V = V − V = Q / C − IR
ab
6 µF
6Ω
6 µF
6 µF
3.00 µF
= (3.6 × 10 −5 C)/(6.0 × 10-6 F) - (2.0 A)(6.0 Ω) = -6.00 V.
b) Point b is at the higher potential.
c) If the switch is closed: Vb = Va = (2.00 A)(3.00 Ω) = 6.00 V.
d) New charges are: Q3 = CV = (3.00 ×10 −6 F)(6.0 V) = 1.80 ×10-5 C.
Q6 = CV = (6.00 × 10 −6 F)(-12.0 V) = -7.20 × 10-5 C.
→ ∆Q3 = +3.60 × 10 −5 C - (1.80 × 10-5 C) = +1.80 × 10-5 C.
∆Q6 = −3.60 × 10 −5 C - (-7.20 ×10 -5 C) = +3.60 ×10 -5 C.
The total charge flowing
through the switch is
5.40 x 10-5 C.