5/27/2011
What is the equivalent resistance between points 1 and 2.
Phy2049
•Circuits have capacitors and resistors.
•Capacitors in parallel
Ceq = C1 + C2
1
in series
= 1
+ 1
C
C
C
1
•Resistors in parallel
= 1
+ 1
R
R
R
in Series
Req = R1 + R2
eq
1
2
eq
1
2
A. 5 Ω
B. 7 Ω
C. 2 Ω
D. 11Ω
E. NOT
You measure the current through the 4 Ω wire and find it to be 3A.
What is the potential difference between points 1 and 2?
•Kirchhof’s two rules: voltage around the
loop and currents at a junction.
•P = iV = i2R = v2/R
Ei−iR = 0
A. 12 V
B. 30V
C. 15V
D. 20V
E. NOT
Multiloop circuits :
Current in a single loop circuit
Consider the circuit shown in the figure. We assume
Consider the circuit shown in the figure. There
are three brances in it: Branch bad, bcd, and bd.
that the emf device is ideal and that the connecting
We assign currents for each branch and define the
current directions arbitrarily. The method is self
correcting. If we have made a mistake in the direction
wires have negligible resistance. A current i flows
through the circuit in the clockwise direction.
of a particular current the calculation will yield
a negative value and thus provide us with a warning.
In a time interval dt a charge dq = idt passes through the circuit. The battery is
doing work dW = Edq = Eiidt.i Using energy conservation we can set this amount
of work equal to the rate at which heat is generated on R. Eiidt = Ri 2 dt →
Ei= Ri → Ei−iR = 0
Kirchhoff put the equation above in the form of a rule known as Kirchhoff's loop rule
(KLR for short)
KLR : The algerbraic sum of the changes in potential encountered in a complete
traversal of any loop in a circuit is equal to zero.
We asign current i1 for branch bad, current i2 for branch bcd, and current i3
for branch bd. Consider junction d. Currents i1 and i3 arrive, while i2 leaves.
Charge is conserved thus we have: i1 + i3 = i2 . This equation can be fomulated
as a more general principle knwon as Kirchhoff's junction rule (KJR)
KJR : The sum of the currents entering any junction is equal to the sum of the currents
leaving the junction
The rules that give us algebraic sign of the charges in potential through a resistor
and a battery are given on the next page.
R
i
Resistance Rule :
∆V = -iR
R
in the direction of the current, the change in the
∆V = +iR
motion
-
potential ∆V = −iR
For a move through a resistance in the direction
motion
∆V = +E
For a move through an ideal emf device
in the direction of the emf arrow,
For a move through an ideal emf device in a direction
motion
Kirchhof’s rules:
Choose current (magnitude and direction) in
all branches. At junction a (or b), I1 = I2+I3
potential ∆V = +iR
the change in the potential ∆V = +E
+
Size and direction of current in E1and E2
Va-Vb
opposite to that of the current, the change in the
EMF Rule :
+
R1 = 1 and R2 = 2 Ὠ
E1 = 2, E2 = E3 = 4 V
For a move through a resistance
motion
i
(27 – 11)
(27 – 5)
∆V = -E
opposite to that of the emf arrow,
the change in the potential ∆V = −E
(27 – 6)
. (a) We note that the R1 resistors occur in series
pairs, contributing net resistance 2R1 in each
branch where they appear. Since ε2 = ε3 and R2 =
2R1, from symmetry we know that the currents
through ε2 and ε3 are the same: i2 = i3 = i.
Therefore, the current through ε1 is i1 = 2i. Then
from Vb – Va = ε2 – iR2 = ε1 + (2R1)(2i) we get
i=
ε 2 − ε1
4.0 V − 2.0 V
=
= 0.33 A.
4R1 + R2 4 (1.0 Ω ) + 2.0 Ω
Therefore, the current through ε1 is i1 = 2i = 0.67 A.
(b) The direction of i1 is downward.
(c) The current through ε2 is i2 = 0.33 A.
(d) The direction of i2 is upward.
(e) From part (a), we have i3 = i2 = 0.33 A.
(f) The direction of i3 is also upward.
(g) Va – Vb = –iR2 + ε2 = –(0.333 A)(2.0 Ω) + 4.0 V = 3.3 V.
Consider the right box loop, go clockwise.
I3(2R1)-E3+E2-I2R2 = 0 I3 = I 2 for the
numbers above. And I1 = 2I2
Now the left loop, clockwise:
2 + 2I1 +2I2 - 4 = 0 I2 = 0.33 A = I3
I1 = 0.67 A
The current directions are correct as chosen
at the outset.
1
5/27/2011
R = 4 Ω, V = 4 V, what is the current through R?
R = 4 Ω, V = 4 V, what is the current through R?
4+4+4-4+4I=0
Obtain current in all branches of this circuit by repeated application of
Kirchhof’s loop and junction rules
j
I = -2A
• What is i? All V = 10 Volts and all R = 4 Ohms
j
k
k
• What is i? All V = 10 Volts and all R = 4 Ohms?
• What is j? What is k?
• What is i? All V = 10 Volts and all R = 4 Ohms
-10 +10+10-10+10-10-10-10-10-10 +10 I = 0 I = 4A
2
5/27/2011
• Three circuits are
connected to a
battery. Rank them
by the
– final charge
– Time constant
3