DC circuits
Dr Jacob Dunningham
School of Physics and Astronomy University of Leeds
EM-L6-1
Review of Lecture 5
• Parallel capacitors
C=
X
Ci
i
• Series capacitors
X 1
1
=
C
i Ci
• Dielectric
E=E0 / κ ; =κ · 0
• Current
I = Q / t ; I = q · n · A · vd
Review
EM-L6-2
Overview
The plan for todays lecture
• Resistor circuits
• Kirchhoff’s Rules
• Summary
Review
EM-L6-3
Resistor circuits
EM-L6-4
Resistivity
Ohms law: for many materials R is constant
R=V /I
Resistance of a wire depends on length L and cross section A.
The resistivity ρ is a characteristic of the conducting material.
R=ρ·
Resistor circuits
L
A
EM-L6-5
Power dissipated
Rate of energy loss
∆U
∆Q · V
=
−
∆t
∆t
P = I ·V
Or using Ohms’s law V = R · I this can be rewritten
P = I · V = I 2 · R = V 2/R
Resistor circuits
EM-L6-6
Series resistors
Current the same in both resistors
V = I R1 + I R2 = I (R1 + R2)
Equivalent resistance
Req = R1 + R2
For several resistors in series
R=
X
Ri
i
Resistor circuits
EM-L6-7
Parallel resistors
The voltage is the same across both resistors
V = I1 · R1 = I2 · R2
The equivalent resistance is
1
I
I + I2
V /R1 + V /R2
1
1
=
= 1
=
=
+
Req
V
V
V
R1
R2
For several resistors in parallel
X 1
1
=
R
i Ri
Resistor circuits
EM-L6-8
Example: equivalent resistance
Find the equivalent resistance and the total current
Resistor circuits
EM-L6-9
Example: equivalent resistance
The resistance of the parallel component is
1
Rp = 1
= 4Ω
1
12 + 6
So the total resistance is Rt = 2 + 4 = 6Ω.
And the total current is: I = V /R = 18/6 = 3A.
http://phet-web.colorado.edu
Resistor circuits
EM-L6-10
Kirchhoff’s rules
Kirchhoff’s rules apply to any circuit
• When any closed-circuit loop is traversed, the algebraic sum
of the changes in potential must equal zero.
• At any junction point in a circuit where the current can divide,
the sum of the currents into the junction must equal the sum
of the currents out of the junction.
Kirchhoff’s Rules
EM-L6-11
Example 1: Kirchhoff’s rules
Taking a clockwise loop starting at point ‘a’
−IR1 − IR2 − E2 − Ir2 − IR3 + E1 − Ir1 = 0
Solving for the current, I, we obtain
I=
E1 − E2
R1 + R2 + R3 + r 1 + r 2
(Note: For E2 > E1 , I < 0, i.e. the current is anticlockwise)
Kirchhoff’s Rules
EM-L6-12
Example 2: Kirchhoff’s rules
(a) Find the current in each branch of the circuit
(b) Find the energy dissipated in the 4Ω resistor in 3 s
Kirchhoff’s Rules
EM-L6-13
Example 2: Kirchhoff’s rules
1. Apply junction rule to point b:
2. Apply loop rule to outer loop:
i.e.
I = I1 + I2
12−2I2 −5−3(I1 +I2) = 0
7 − 3I1 − 5I2 = 0
3. For the third condition, apply the loop rule to bcdeb:
i.e.
−5 + 4I1 − 2I2 = 0
Kirchhoff’s Rules
EM-L6-14
Example 2: Kirchhoff’s rules
Solve equations 2 and 3:
7 − 3I1 − 5I2 = 0
−5 + 4I1 − 2I2 = 0
This gives:
I1 = 1.5 A
From equation 1, we get:
I2 = 0.5 A
I = I1 + I2 = 2.0 A
Finally, the power in the 4Ω resistor is:
The energy dissipated in 3 s is:
Kirchhoff’s Rules
P = I12R = 9 W
W = P ∆t = 9 × 3 = 27 J
EM-L6-15
Summary
• Resistors in series or in parallel
R=
X
i
Ri
X 1
1
=
R
i Ri
• Power dissipated
P =V ·I
• Kirchhoff’s rules
Reading: Tipler, sections 25-3, 25-4, 25-5
Preparation: Tipler, sections 25-6, 26-1
Summary
EM-L6-16