1. Electric Circuits Assessment Problems
Chapter 4
Circuit Elements

2. 4.13 Find the power delivered by the 2 A current source in the circuit shown.
Ans 𝑉𝑎− 𝑉𝑎−25 10 =2 Va=35V P2A= 35V(2A)=70W # Va

3. 4.14 Find the power delivered by the 4 A current source in the circuit shown.
Ans: 30Ix= 𝑉1−𝑉2 4 (30) =7.5V1-7.5V2 𝑉1− 𝑉1−𝑉 =0 𝑉2−𝑉1 4 + 𝑉2 6 + 𝑉2−𝑉3 3 =0 𝑉3−𝑉 𝑉3−7.5𝑉1+7.5𝑉2 5 =0 V2 V1 V3

4. 3V1-V2=240 -3V1+9V2-4V3= V1+17.5V2+8V3=60 V2=3V V1-4V3= V1+8V3= V1-8V3=4320 V1=110V V2=90V V3=120V P4A=(V3-V1)(4)=( )(4)=40W #

5. b) How much power does the 120 V source deliver to the circuit?
4.15
a) Use a series of source transformations to find the voltage v in the circuit shown.
b) How much power does the 120 V source deliver to the circuit?
Ans:
(a)
Rth=1.6+(6//5//20) =4Ω
𝐸𝑡ℎ 𝐸𝑡ℎ 𝐸𝑡ℎ− =0
Eth=72V
V= =48V#
Rt=6//5//20=2.4Ω
It= =30A
V=[30 ( )](8)=48V #
Eth 6A
12A
36A
1.6Ω 8Ω
20Ω 5Ω 6Ω

6. (b) 𝐸𝑡ℎ 6 -36+ 𝐸𝑡ℎ+60 5 + 𝐸𝑡ℎ−120 20 + 𝐸𝑡ℎ 9. 6 =0 Eth=57
(b) 𝐸𝑡ℎ 𝐸𝑡ℎ 𝐸𝑡ℎ− 𝐸𝑡ℎ 9.6 =0 Eth=57.6V I1= 120− =3.12A P120V=3.12 (120) =374.4W #
Eth i1

7. 4.16 Find the Thevenin equivalent circuit with respect to the terminals a,b for the circuit shown.
ANS:
Rth=[(5//20)+8]//12
=6Ω #
𝑉𝑎− 𝑉𝑎 20 + 𝑉𝑎−𝑉𝑡ℎ 8 =0
𝑉𝑡ℎ−𝑉𝑎 𝑉𝑡ℎ− =0
Vth=64.8V # Va
Vth

8. 4.17
Find the Norton equivalent circuit with respect to the terminals a,b for the circuit shown.
Ans: Rth=12//(2+8+10) =7.5Ω # Eth= =45V In= = 6A # In= =6A #

9. 4.18
A voltmeter with an internal resistance of 100 kΩ is used to measure the voltage vAB in the circuit shown. What is the voltmeter reading?
Eth
ANS
Rth=15k+(60k//12k)
=25kΩ
𝐸𝑡ℎ 60𝑘 - 18m + 𝐸𝑡ℎ+36 12𝑘 =0
Eth=150V
VAB= 𝑘 25𝑘+100𝑘 = 120V #

10. 4.19
Find the Thevenin equivalent circuit with respect to the terminals a,b for the circuit shown.
Ans:
Ix= 𝐸𝑡ℎ 8
3𝐸𝑡ℎ 𝐸𝑡ℎ− 𝐸𝑡ℎ 8 =0
Eth=8V #
It=ix+3ix+ 𝐸𝑡ℎ 2
It=1+3+4=8A
Rth= 8 8 =1Ω #
Eth
Eth it

11. 4.20
Find the Thevenin equivalent circuit with respect to the terminals a,b for the circuit shown. (Hint: Define the voltage at the leftmost node as v, and write two nodal equations with VTh as the right node voltage.)
Ans:
IΔ= 𝐸𝑡ℎ 40
v 60 + v − 160 𝐸𝑡ℎ 40 − E𝑡ℎ 20 =4
𝐸𝑡ℎ+160 𝐸𝑡ℎ 40 −𝑣 𝐸𝑡ℎ 𝐸𝑡ℎ 40 =0
Eth=30V #
V=172.5V
Eth V

12. It= =3A Rth= 30 3 =10Ω #
Eth it

13. 4.21
a) Find the value of R that enables the circuit shown to deliver maximum power to the terminals a,b.
b) Find the maximum power delivered to R.
Ans:
(a)
Vφ=Va-20
𝑉𝑎− 𝑉𝑎− V𝑎−𝐸𝑡ℎ 4 =0
𝐸𝑡ℎ−𝑉𝑎 4 + 𝐸𝑡ℎ−𝑉𝑎+20−100 4 =0
Eth=120V #
Va=80V Va
Eth

14. Vφ= =40V It= −40 4 = 40A Rth= 𝐸𝑡ℎ 𝐼𝑡 = =3Ω # (b) PRmax= 𝐸𝑡ℎ2 4𝑅𝑡ℎ = (3) =1200W=1.2kW # Vφ 4 - + 2 4
Eth Vφ it b a

15. Assume that the circuit in Assessment
4.22
Assume that the circuit in Assessment
Problem 4.21 is delivering maximum power to the load resistor R.
a) How much power is the 100 V source delivering to the network?
b) Repeat (a) for the dependent voltage source.
c) What percentage of the total power generated by these two sources is delivered to the load resistor
Ans 𝑉𝑎− 𝑉𝑎− 𝑉𝑎−𝑉𝑏 4 =0 𝑉𝑏−𝑉𝑎 4 + 𝑉𝑏 3 + 𝑉𝑏−𝑉𝑎+20−100 4 =0 Va=60V Vb=60V I1= 100−60 4 =10A I2= −60 4 =20A 3Ω Va Vb i2 i1

16. PR Pdeliver = 1200 3800 = 0.31578 =31.58% # (a) P100V=30(100)=3000W #
(b)PVφ=20(40)=800W #
(c)Pdeliver= =3800W
PR = =1200W
PR Pdeliver = = =31.58% #