PHYSICS 3050
Total - 38 marks
Answers to Problem Set 1
was due October 7, 2010
”The wireless music box has no imaginable commercial value. Who would pay
for a message sent to nobody in particular?”
David Sarnoff’s associates in response to his urgings for investment in the radio in the 1920s.
• For full solutions to 1–3 and 5–7 see scanned solutions.
1. Given R1 = 5 kΩ, R2 = 4 kΩ, R3 = 2 kΩ, V1 = 10 V and V2 = 6 V, find i1 and i2 .(4
marks)
• i1 = 18/19 mA = 0.947 mA and i2 = 25/19 mA = 1.316 mA.
2. Variation on Text Problem 33, page 46 (4 marks). Do the same problem but assume
the battery supplies 19 V instead of 6 V.
(a) 5 A through the 3 Ω, 1 A through the 4 Ω, and 2 A through each 2 Ω resistor.
(b) 15 V across the 3 Ω and the other three are in parallel so there is 19 V - 15 V =
4 V across each of them.
(c) P = 95 W.
3. Variation on Text Problem 34, page 46 (4 marks). Use any method you want to
calculate the current through and the voltage drop across (a) the 10 Ω resistor and
(b)the 2 Ω resistor.
(a) 1 A and 10 V.
(b) -2 A and -4 V
4. 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? (4 marks)
• 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.
5. Calculate the Thévenin circuit ‘as seen by’ RL for the circuit of Fig 1.33 (pg 47) of the
text. (4 marks)
• VEQ = 9 V and REQ = 12.5 Ω.
6. Obtain the Thévenin and Norton equivalent circuits ‘as seen by’ the 10 Ω resistor in
the circuit of Fig 1.30 (pg 46) of the text. (4 marks)
• VEQ = 12 V, REQ = 2 Ω, and iEQ = 6 A.
7. Variation on Text Problem 39, page 47.
(a) Find the Thévenin equivalent circuit ‘as seen by’ the terminals x−y for the circuit
of Fig 1.32 but assume the voltage source supplies 10 V and the 30 Ω resistor is
replaced by a 20 Ω resistor. (4 marks)
• VEQ = 10 V and REQ = 40/6 Ω.
(b) What is the current through the 40 Ω resistor? (4 marks)
• i = 0.25 A
8. (6 marks) Find the Thévenin and Norton equivalent circuits for the following circuit:
R1 R2
R1 + R2
(I2 R1 + V1 )R2
− I1 R2
R1 + R2
[I2 R1 + V1 − I1 (R1 + R2 )]R2
R1 + R2
[V1 − I1 R2 + R1 (I2 − I1 )]R2
R1 + R2
V1 − I1R2 + R1 (I2 − I1 )
2R1 + R2
REQ = R2 +
VEQ =
=
=
IEQ =