1. Electric Circuits Assessment Problems
Chapter 2
Circuit Elements

2. 2.1 For the circuit shown, A) What value of vg is required in order for the interconnection to be valid? b) For this value of vg find the power associated with the 8 A source.
2.1 Ans. (a) (b) ib = −8A P8A =IV=(8)(-2) vg = ib 4 = −8 4 =-2v # =-16W(提供) #

3. 2.2 For the circuit shown, a) What value of a is required in order for the interconnection to be valid? b) For the value of a calculated in part (a), find the power associated with the 25 V source.
2.2 Ans. (a)αVx = −15A , α(-25)=-15 α=0.6A/V # (b)P25V=IV=(-25)(-15)=+375(消耗) #

4. 2.3 For the circuit shown,
a) If vg = 1 kV and ig = 5 mA, find the value of R and the power absorbed by the resistor.
b) If ig = 75 mA and the power delivered by the voltage source is 3 W, find vg, R, and the power absorbed by the resistor.
c) If R= 300Ω and the power absorbed by R is 480 mW, find Ig and vg.
Ans:
(a)R= 𝑉 𝐼 = 1𝑘𝑉 5𝑚𝐴 =200kΩ #
P=IV=(5mA)(1kV)=5W #
(b)Pvg=IgVg ; 3=75m(Vg) ; Vg=40V #
R= 𝑉 𝐼 = 40 75𝑚 =533.33Ω # PR= 75m(40V)=3w #

5. (C) PR= V2 R ; 480m= 𝑉𝑔2 300 ; 𝑉𝑔 =12V # Ig= V𝑔 R ; Ig= 12 300 = 40mA #

6. 2.4 For the circuit shown,
a) If ig = 0.5 A and G = 50 mS, find vg and the power delivered by the current source.
b) If vg = 15 V and the power delivered to the conductor is 9 W, find the conductance G and the source current ig.
c) If G = 200 μS and the power delivered to the conductance is 8 W, find ig and vg.
Ans: (a)Vg = 𝐼𝑔 𝐺 = 𝑚𝑠 =10V # Pig=IgVg=(-0.5)(10)=-5W(提供) # (b)P= 𝑉2 𝑅 ;9= 152 𝑅 , R=25Ω；G= 1 𝑅 , G= 1 25 =40mS # ig= 𝑉 𝑅 = =0.6A #

7. (C)
P=GVg2;8=(200μS)V2 ; Vg=200V #
Ig=GVg=(200μ)(200V)=40mA #

8. 2.5
For the circuit shown, calculate (a) i5; (b) v1; (c) v2; (d) v5; and (e) the power delivered by the 24 V source.
Ans:
(a)i5= 𝑉 𝑅𝑇 = =2A #
(b)V1=(i5)(2Ω)=(-2)(2)=-4V #
(c)V2=(i5)(3Ω)=(2)(3) = 6V #
(d)V5=(i5)(7Ω)=(2)(7) =14V #
(e)P24V=(i5)(24V)=(-2)(24)=-48W #

9. 2.6
Use Ohm’s law and Kirchhoff’s laws to find the value of R in the circuit shown.
Ans:
VR = =80V
80 𝑅 =
80 𝑅 = 5+15
R = 4Ω #

10. 2.7 Ans: (a) (b) 𝑉−25 𝑅 =0 it = 25 100+25 𝑉−15 𝑅 =0.1 =0.2A
a) The terminal voltage and terminal current were measured on the device shown. The values of vt and i, are provided in the table.Using these values, create the straight line plot of vt versus it. Compute the equation of the line and use the equation to construct a circuit model for the device using an ideal voltage source and a resistor.
b) Use the model constructed in (a) to predict the power that the device will deliver to a 25 Ω resistor.
Ans:
(a) (b)
𝑉−25 𝑅 =0 it =
𝑉−15 𝑅 = =0.2A
𝑉−5 𝑅 = P25Ω=0.22(25)
𝑉−0 𝑅 = =1W #
V=25V R=100Ω #

12. 2.8
Repeat Assessment Problem 2.7 but use the equation of the graphed line to construct a circuit model containing an ideal current source and a resistor.
Ans:
(a)
I = 𝑉 𝑅 = = 0.25A#
R=100Ω #
(b)
I25Ω= =0.2A
P25Ω=0.22 (25) = 1W #

13. 2.9
For the circuit shown find
(a) the current i1 in microamperes,
(b) the voltage v in volts,
(c) The total power generated, and
(d) the total power absorbed.
Ans:
(a) 6−𝑉𝐴 54𝑘 + 30( 6−𝑉𝐴 54𝑘 ) = 𝑉𝐴 6𝑘 ；i1= 6−𝑉𝐴 54𝑘
VA= 4.65V
i1=25μA #
(b) V=8-1.8k(0.75m)-4.65=2(正負相反)
V=-2V #

14. 5V -IV -(25μ)(5) -125μW 1V -(25μ)(1) -25μW 8V -(750μ)(8) -6000μW 54kΩ
提供功率
(c) 5V
-IV
-(25μ)(5)
-125μW 1V
-(25μ)(1)
-25μW 8V
-(750μ)(8)
-6000μW
-125μW+(-25μW)+(-6000μW) = -6150μW #
消耗功率
(d)
54kΩ
I2R
(25μ) 2(54k)
33.75μW
6kΩ
(775μ) 2(6k) μW
1.8kΩ
(750μ) 2(1.8k)
1012.5μW
30i1
-IV
-(750μ)(-2)
1500μW
33.75μW+( μW)+(1012.5μW) )+(1500μW) = 6150μW #

15. 2.10
The current iΦ in the circuit shown is 2 A. Calculate
a) vs,
b) the power absorbed by the independent voltage source,
c) the power delivered by the independent current source,
d) the power delivered by the controlled current source,
e) the total power dissipated in the two resistors.

16. 5A 2A 3A 4A
-1A Va
(a)
(c)
P5A= -(5A) (60V) = -300W(提供) #
(d)
P2iΦ = -(4A) (10V) = -40W(提供) #
(e)
P10Ω = (1A)2 (10Ω) = 10W(消耗)
P30Ω = (2A)2 (30Ω) = 120W(消耗)
Ptotal= 10w + 120w = 130W #
Va = 2A (30Ω)= 60V
60-Vs= -1A (10Ω)
Vs=70V #
(b)
PVs= 3A (70V) = 210W(消耗) #