PHYSICS 3050
Total - 30 marks
Solutions to Problem Set 1
was due October 1, 2009
“Computers are useless. They can only give you answers.”
Pablo Picasso (1881 - 1973)
1. (4 marks) Given R1 = 5 kΩ, R2 = 4 kΩ, R3 = 2 kΩ, V1 = 10 V and V2 = 6 V, find i1
and i2 .
i1
V2
R1
V1
i2
R3
R2
• i1 = 18/19 mA = 0.947 mA and i2 = 25/19 mA = 1.316 mA.
2. (4 marks) Calculate i1 , i2 , and i3 .
i1
1 kΩ
i2
1 kΩ
1 kΩ i 3
12 V
1 kΩ
2 kΩ
• i1 = 72/13 mA = 5.54 mA, i2 = 132/13 mA = 10.15 mA, and i3 = 12/13 mA = 0.92
mA.
3. (4 marks) You are a given a black box with two output terminals. When a load resistor
of 20 kΩ is connected across the terminals, the voltage across the load is measured to
be 15 V. The voltage across a 50 kΩ load is found to be 18 V. What are the equivalent
circuit values of VEQ , REQ and IEQ for the black box?
• From voltage division we have for VL , the voltage across the load resistor RL ;
VL = VEQ
RL
REQ + RL
Putting in the two values of VL and RL allows to solve for VEQ and REQ (two equations
and two unknowns) giving VEQ =270/13 V = 20.8 V and REQ = 100/13 kΩ = 7.7 kΩ.
The Norton current iEQ is then VEQ /REQ = (270 V)/(100 kΩ) =2.7 mA.
4. Obtain the Thévenin equivalent circuit (i.e., VEQ and REQ ) to the following circuit
considering as output terminals:
(a) AB (3 marks)
• Doing it just as in class, we get VEQ from the case where RL → ∞ and when
RL → 0 we use REQ =VEQ /iAB . So, in the limit RL → ∞ we have VEQ =VAB
which we can just get from voltage division. That is,
VEQ = VAB = (10V + 6V)
(3 kΩ)
= 12 V
(3 kΩ + 1 kΩ)
Further, the limit RL → 0 is equivalent to shorting the 3 kΩ so the current is just
that generated by the total 16 V across the 1 kΩ resistor. That is,
iAB =
VEQ
12 V
16 V
= 16 mA =⇒ REQ =
=
= 0.75 kΩ
1 kΩ
iAB
16 mA
(b) AC (3 marks)
• Now VEQ =VAC which is always 6 V by definition of the emf between points A
and C. If we try to short AC, we then have no resistance across 6 V giving an
infinite current. So REQ = VEQ /iAC = (6 V)/∞ = 0 Ω.
6V
CO
1 kΩ
O
A
3 kΩ
10 V
O
B
5. (6 marks) Given that V1 =12 V, V2 =9 V, R1 =10 kΩ, R2 =20 kΩ, R3 =2 kΩ, and R4 =1
kΩ, what is the the Thévenin equivalent circuit “seen” by RL ? What is the current
through and the voltage across RL for three different values of RL : 100 Ω, 1 kΩ, and
10 kΩ?
R3
R4
V2
V1
R1
RL
R2
• The circuit “as seen by” RL is:
R3
R4
i
V1
A
V2
B
[Note to self – R1 and R2 are irrelevant to the problem since V1 and V2 maintain a
constant voltage across them, respectively.]
Now as I’ve drawn it, VEQ =VAB and, with i so defined, we have from Kirchoff’s voltage
law:
V1 − iR3 − iR4 − V2 = 0
so VAB =V1 -iR3 =V2 +iR4 and
i=
V1 − V 2
(12 V − 9 V)
=
= 1 mA
R3 + R 4
(2 kΩ + 1 kΩ)
And therefore
VEQ = VAB = V1 − iR3 = 12 V − (1 mA)((2 kΩ) = 10 V
[Note: V2 +iR4 = 9 V + (1 mA)(1 kΩ) = 10 V ... as it must.]
To calculate REQ we need to short A and B to get iAB = iEQ , the current through
AB. This current is then the sum of the currents i3 and i4 due to V1 across R3 and
V2 across R4 , respectively. This is shown below.
R3
R4
i3
i4
V1
A
V2
iAB
B
So
iEQ = iAB = i3 + i4 =
and, therefore,
REQ =
V1 V2
9V
12 V
+
=
+
= 15 mA
R3 R4
1 kΩ 2 kΩ
10 V
2
VEQ
=
= kΩ
iEQ
15 mA
3
The Thévenin equivalent circuit with RL put back in is then:
R EQ
iL
VEQ
RL
Now we get trivially that:
10 V
= 10 mA
kΩ + 31 kΩ
10
1
V
= iL RL = (10 mA)( kΩ) =
3
3
iL =
VL
VEQ
=
REQ + RL
2
3
6. (6 marks) Find the Thévenin and Norton equivalent circuits for the following circuit:
A
I1
R1
V1
I2
R2
R2
B
R1 R2
R1 + R 2
(I2 R1 + V1 )R2
− I 1 R2
R1 + R 2
[I2 R1 + V1 − I1 (R1 + R2 )]R2
R1 + R 2
[V1 − I1 R2 + R1 (I2 − I1 )]R2
R1 + R 2
V1 − I1R2 + R1 (I2 − I1 )
2R1 + R2
REQ = R2 +
VEQ =
=
=
IEQ =