电路原理题目

第一章

1.图中所示电路中，已知a点、b点的电位分别为

φa=10V,φb=5V。

则电动势E＝____V:

电压U＝____V:

由参考方向，得

E=φa−φb,U=φa−φb

由已知得

E=10−5=5V,U=10−5=5V

2. 蓄电池端纽a和b两端的电压Uab和电动势Eab的关系是：

Eab=UabEab=−Uab

Eab=−Uab-

由参考方向知道，电压Uab是指电压从a点到b点的降低。电动势Eab是指电压从a点到b点的升高。由此得Eab＝－Uab

3. 图中所示电路中，已知a点、b点的电位分别为

φa=10V,φb=5V。

则电动势E＝____V:

电压U＝____V:

由参考方向，得

E=φb−φa,U=φa−φb

由已知得

E=5−10=−5V,U=10−5=5V

4.

图中所示电路中，已知a点、b点的电位分别为

φa=10V,φb=5V。

则电动势E＝____V:

电压U＝____V:

由参考方向，得

E=φa−φb,U=φb−φa

由已知得

E=10−5=5V,U=5−10=−5V

5. 图中所示电路中，已知a点、b点的电位分别为

φa=10V,φb=5V。

则电动势E＝____V:

电压U＝____V:

由参考方向，得

E=φb−φa,U=φb−φa

由已知得

E=5−10=−5V,U=5−10=−5V

6.二端元件的端电压和流过的电流如图所示。

则二端元件发出的功率P＝____W:

由电流的参考方向与电压降的方向一致,得，发出功率P＝－2×3＝－6W

7. 二端元件的端电压和流过的电流如图所示。

则二端元件发出的功率P＝____W:

由电流的参考方向与电压降的方向不一致,得，发出功率P＝2×3=6W

The reference direction of the voltage U across the resistor R is

designated as: right is positive and left is negative, while the

reference direction of it’s the current I is along the arrow from left

to right, then the relationship of U and I is:

U=RI

U=RI-

U=−RI

Based on Ohm’s law, noting that the reference directions of U and I for the

resistor R are non-associated.

the circuit in the figure,the voltage Uab is:

Uab=−14VUab=−14V-

Uab=11VUab=−5VUab=1V

Mark the reference direction of the voltage U as shown in figure.

We get

U=3∗1=3V

According to KVL, the voltage is:

Uab=−8+3=−5V

the circuit in the figure, the current

I=____A:

Mark the reference direction of the current I1 as shown in figure.

Based on Ohm’s law,

I1=12/2=6A

According to KCL,

I=4−I1=−2A

10. For the circuit in the figure, the current

I=____A and the

voltage

U=____V :

Mark the reference directions of the current I1 and the voltage U1

as shown in figure.

Based on Ohm’s law,

I1=12/2=6A

U1=3∗8=24V

According to KCL and KVL,

I=8−I1=2A

U=U1+12=36V

11. For the circuit in the figure, the current

I =____A and the

voltage

U1=____V，U2 =____V:

Mark the reference directions of the current I1 and the voltage U3

and U4 as shown in figure.

Based on Ohm’s law,

I1=9/1=9A

U3=3∗2=6V

U4=2∗2=4V

According to KCL and KVL,

I=3+2−I1=−4A

U1=U3+9=15V

U2=U4+9=13V

the circuit in the figure, the current

I =____A and the

voltage

U1=____V，U2 =____V:

Mark the reference directions of the current I1 and the voltage U3

and U4 as shown in figure.

Based on Ohm’s law,

I1=9/1=9A

U3=3∗2=6V

U4=2∗2=4V

According to KCL and KVL,

I=3+2−I1=−4A

U1=U3+9=15V

U2=U4+9=13V

13. Given the circuit as shown in the figure,(1) when

I=4A，the

current of current sourceIS ＝______A ;(2) when

U=9V，the

current of current sourceIS ＝______A:

Write the equation applying KVL,

U1=3U1+I×1

According to Ohm’s law and KCL,

U=3×(U1/1+I)

According to Ohm’s law and KCL,

IS=(U+U1)/2+U1/1+I

Combine and solve the equations above by using the known values.

（1）当I = 4A时， Is = 4A（2）当U = 9V时， Is = 6A

14. In the figure,the current of current source

IS= (A)?

Write the equation applying Ohm’s law and KVL,

U=−100U1+40×2=−100U1+80

According to Ohm’s law,

U1=0.2IS

According to KCL and Ohm’s law,

U=5(IS−2)

Combining the above equations, we have

IS=3.6A

15. Refer to the circuit of the figure, the power delivered by the

independent source is

P= _____W.

Write KCL on the upper node,

U/2+U/3+0.5U=12

Thus,

U=9V

The power delivered by the independent source is

Pgen=12×9=108W

16. In the following circuit, find the node voltage

V3=__8.75____V. (hint: use simulation tools.)

第二章

1. As Fig2-1 shown, the input resistance of the two circuits

are______Ω and______Ω

Simplify the circuit, we get

The bridge is balanced and the 5 ohm resistor can be removed, we

get

From the series-parallel formula, the solved resistance is

R=10Ω

Solution for (b)：

Simplify the circuit, we get

From the series-parallel formula, the solved resistance is

R=10Ω

2. The circuit in Fig 2-2 can be simplified as：Fig（a）, Fig（b）, Fig（c）, Fig（d）. Choose （ ）or（ ）.

解析：In the Fig.2-2, the 3 ohm resistor connected in series with the

5A current source doesn’t contribute to the port voltage, it only

affects the voltage of current source. So it can be removed in the

simplification.

Using source transformation to the circuit in Fig above, we get the

circuit shown in Fig below

3. The circuit in Fig 2-3 can be simplified as：Fig（a）, Fig（b）, Fig（c）, Fig（d）. Choose （ ）.

&：Remove the 10 ohm resistor connected in parallel with the 6V

voltage source, because it doesn’t contributes to the port voltage,

then we get

Transform the 2A current source in parallel with the 10 ohm resistor

into a voltage source in series with a resistor, then we get

It’s finally simplified as

4.

Using source transformation to solve the current

I(/A) of the circuit

in the figure:

Using source transformation to the 8V voltage source in series with

the 4 ohm resistor, we get

It’s simplified as

解析：Using source transformation to the 6A current source in

parallel with the 2 ohm resistor and the 3A current source in parallel

with the 3 ohm resistor respectively, then we get

The current

I is solved

I=(12−9)/(2+3+5)=0.3A

5. Using source transformation to solve the current

I (/A) of the

circuit in the figure:

解析：To simplify the circuit, remove the 10A current source in

parallel with the voltage source and the 10 ohm resistor in series

with the 5A current source, then we get

Transform the two current voltage in parallel with resistor into

voltage source in series with resistor, and we get

It’s simplified as

Transform the 28V voltage source in series with the 4 ohm resistor

into a 7A voltage source in parallel with a 4 ohm resistor, we get

Two 4 omega resistors in parallel is 2 ohm resistor, and transform

the 7A current source in parallel with the 2 ohm resistor into a 14V

voltage source in series with a 2 ohm resistor, then we get

The current I is solved

I=(14−25)/(2+3+5)=−1.1A

6. The voltage

Uab in the circuit shown in the figure is______V:

解析：Using source transformation to the dependent source in the

circuit, we get

Applying the KCL and KVL, the equation is written as

4(1−I)+4I=8I

The current is solved

I=0.5A

Based on Ohm’s Law, the voltage is

Uab=8I=0.5×8=4V

7. The voltage

U,

U2 and the power delivered by the current source

in the circuit shown in the figure

isU=______V,U2=______V,Pdeliver=______W:

解析：The 6 ohm resistor in parallel with the 3 ohm resistor can be

simplified as a 2 ohm resistor, then we get

The bridge consists of the right five resistors is balanced, so the

circuit can be simplified further as

U is solved

U=2×(3//2)=2×1.2=2.4V

Based on the current division formula,

U2=U/2=1.2V

The power delivered by the current source is

Pdeliver=2U=2×2.4=4.8W

8. Solve the power delivered by the voltage source and current

source respectively in the circuit shown in the figure,

The power delivered by the 4A current source is ______W:

The power delivered by the 30V voltage source is ______W:

解析：Step 1:

Suppose the voltage of the 3 ohm resistor is U1 , the current

through the 6 ohm resistor is I1 , the reference direction is shown in

the Fig

Step 2:

The 20 ohm resistor in series with the 4A current source and the 10

ohm resistor in parallel with the 30V voltage source don’t contribute

to U1 and I1 . Remove them when solve U1 and I1 , the equivalent

circuit is shown in Fig below

Step 3:

Using source transformation, then we get

Step 4:

U1 is solved

U1=(4+5)×(3//6//6)=9×1.5=13.5V

Step 5:

Solve I1 in Fig of step 2

I1=(−30+U1)/6=−2.75A

Step 6:

Suppose the voltage of current source is U, the current through the

voltage source is I in Fig , the reference direction is shown in Fig

below：

Step 7:

Solve the circuit in Fig above, we get

U=4×20+U1=80+13.5=93.5V

I=30/10−I1=3−(−2.75)=5.75A

Step 8:

Calculation

The power delivered by the voltage source is

Pdeliver=30I=30×5.75=172.5W

The power delivered by the current source is

Pdeliver=4I=4×93.5=374W

9. (Simulation) The balanced bridge is also quite useful in control

systems. Let us take the temperature control system as an

example. Suppose we have a thermistor whose resistance R is in

direct proportion to the environment temperature. That means

when the temperature increase ΔT centidegree, the resistance of

the thermistor will raise ΔR ohm accordingly. The V in the circle in

the following Figure is a voltmeter with very large resistance. It is

clear that the voltmeter will have zero value when ΔR=0.

If the environmental temperature causes the thermistor changing

its resistance to R+ΔR, the bridge is not balanced anymore. Deduce

the expression of voltmeter’s voltage, find the voltage using any of

your simulation software, and fill in the blank

uV=____2.5____mV

with

uS=10V, R=10kΩ, and ΔR=10Ω:

第三周

1. As the circuit shown in figure 5-1,the voltage

source

US1=12V,US2=8V,the internal resistance

RS1=4Ω,RS2=4Ω;

the resistance

R1=20Ω,R2=40Ω,R3=28Ω,R4=8Ω,R5=16Ω.Find each

branch current by using loop currents analysis.

I1=_____A,I2=_____A,I3=_____A,I4=_____A,I5=_____A,I6=_____A.

解析：Step 1

Choose three meshes in original circuit as a set of independent

loops, and set loop

currents are Il1,Il2,Il3, as shown in figure (a).

Step 2

Write KVL equations for each loop, we have

(R1+RS1+R4)Il1−R4Il3=US1

(RS2+R3+R5)Il2−R5Il3=US2

−R4Il1−R5Il2+(R4+R5+R2)Il3=0

Step 3

Substitute known values into equations，thus

32Il1−8Il3=12

48Il2−16Il3=8

64Il3−8Il1−16Il2=0

Step 4

We get

Il1=0.4A

Il2=0.2A

Il3=0.1A

Step 5

Get each branch current in terms of loop currents

I1=Il1=0.4AI2=Il2=0.2AI3=Il3=0.1AI4=Il1−Il3=0.3AI5=Il3−Il2=−0.1AI6=Il1−Il2=0.2A

2. ind the voltage

U across the resistor

4Ω by using loop currents

analysis for the circuit as shown in figure 5-2.

U=______V

解析：Assume mesh currents as I1,I2,I3,I4, as shown in figure

(a).

Write KVL equations for each mesh, we have

I1=2

−6I1+18I2−4I3−6I4=0

I3=−3

−6I2−3I3+9I4=−12

Step 2

Rewrite these equations as

18I2−4I3=−6

−6I2−3I3=15

We get

I2=−1A,I3=−3A

Step 4

And the voltage is

U=4(I2−I3)=8V

3. Solve the power delivered by 1A current source by using loop

currents analysis for the circuit as shown in figure 5-3.

Pout=______W

解析；Step 1

Choose a set of independent loops as shown in the figure, and

consider dependent source as independent source, we can get KVL

equations for loops as follow,

(20+30+50)Il1−20Il2−30Il3=0

Il2=1A

Il3=−0.02U

Step 2

Find the relationship between control value and loop current

U=20(Il2−Il1)+0.4U

Rewrite the equation as

U=1003(Il2−I1l)

Step 3

Substitute equation (2) into equation (1) ,we get

U=33.3V

Step 4

The power delivered by 1A current source is

Pout=1×U=33.3W

4. Solve the voltage U across 1A current source and branch

currents I1, I2, I3 by using node voltage analysis for the circuit as

shown in Figure 5-4.

U=_____V,I1=_____A,I2=_____A,I3=_____A.

Step 1

Select a reference node, and designate the voltage of

independent node as Ua, see Figure (a). Write KVL equation for

the independent node

(1/13.7+1/17.2+1/11.9)Ua=5/13.7+15/17.2+10/11.9+1

Step 2

We can get the voltage of independent node is

Ua=14.31V

Step 3

The voltage across the independent current source is

U=Ua+1×5.5=19.8V

Step 4

Branch currents are

I1=Ua−5/13.7=0.680A

I2=Ua−15/17.2=−0.0401A

I3=Ua−1011.9=0.362A

the current I flows through 8V voltage source and the

voltage U across 1.5A current source by using node voltage analysis

for the circuit as shown in figure 5-5.

I=_____A,U=_____V.

解析：In original circuit, designate the reference node and

independent nodes, as shown in figure (a).Write KVL equations for

nodes a and b

Rewrite these equations as

Step 2

Solve equations, we get

Ua=10.75VUb=4.031V

Step 3

The current is

Step 4

The voltage cross the current source is

U=Ua+4×1.5=10.75+6=16.8V

6. Write the node voltage equations for the circuit shown in Figure

5-6, and solve the current I flows through

5Ω resistor.

I=____A

Step 1

Consider dependent source as independent source, write KVL

equations for nodes ① and ②, we have

Step 2

Find the relationship between control value and node voltage

Step 3

Substitute equation (2) into equation (1), we get

Step 4

Thus

U1=−31.5V,U2=30V,I=−6A

the ideal op-amp circuit shown in the figure, the voltage

source

Ua=5V,

Ub=3V, resistors

R1=6kΩ and

R2=12kΩ.Determine

the value of the op-amp’s output voltage

Uo.

解析：Designate the inverting input voltage as U－ and the

noninverting input voltage as U+.

According to the character of “virtual open circuit” for the ideal op-amp, the noninverting input voltage

According to the character of “virtual short circuit” for the ideal op-amp, we obtain

the inverting input voltage U－= U+ = 2V

Write KCL equation on the inverting input terminal

Substitute values into it，thus

8. For the ideal op-amp circuit shown in the figure, the voltage

source

Ua=5V,

Ub=3V, resistors

R1=6kΩ and

R2=12kΩ.Determine

the value of the op-amp’s output voltage

Uo (V).

解析：Designate the inverting input voltage as U－ and the

noninverting input voltage as U+.

According to the character of “virtual open circuit” for the ideal op-amp, the noninverting input voltage

According to the character of “virtual short circuit” for the ideal op-amp, we obtain

the inverting input voltage U－= U+ = 2V

Write KCL equation on the inverting input terminal

Substitute values into it，thus

9. As the circuit shown in the figure, the voltage source

us(t)=3cos4t

V. Solve the current

i=_____mA，when

t=1s.

According to the character of “virtual short circuit”, we have

U−=U+=us(t)=3cos4t

V

According to the character of “virtual open circuit”, we get

Voltage division gives

Hence,

i(t)=cos4t

mA

when t=1s,i=-0.65mA

10.

As the ideal op-amp circuit shown in the figure, determine the

circuit voltage gain Uo/Us.

According to the character of “virtual short circuit”, we have

UA=0

Write KCL equation on node A, we get

Combining the above equations, we have

Hence,

Uo/US=−2

11.

As the ideal op-amp circuit shown in the figure. Select its input

resistance

Rin is：

According to characters of “virtual open circuit” and “virtual short circuit”, we

have

According to the character of “virtual open circuit” for the ideal op-amp, the

input current

Reorganize equations and get

Input resistance

12. As the circuit shown in the figure, the input

voltage

ui=0.2sin314tV, resistance

R1=R2=R3=R4=Ro=RL=10kΩ.

The load current

iL==______mA, when

t=0.01s.

Suppose

ua=u, according to the character of “virtual open circuit”

for the ideal op-amp, the potential of node b

ub=2u

According to the character of “virtual short circuit”, we have

uc=uo

According to characters of “virtual open circuit” and “virtual short

circuit”, write KCL equation on the noninverting input terminal, we

get

Combining the above equations, we have

Reorganize equations and get

u=0.5(ui+uo)

Applying Ohm’s law to the output resistor Ro, and based on the

character of “virtual open circuit” for the ideal op-amp,

Substitute known values into it，thus

the Figure, the u-i relationship of element X isi=Aeu/B ,

wherein A and B are all constants. Analyze the relationship between

the output voltage

uo and the input voltage

ui , and select the

operation function of the circuit.

According to the character of “virtual short circuit”, we have

u+uo=0

that is

u=−uo

According to characters of “virtual open circuit” and “virtual short

circuit”, write KCL equation on the inverting input terminal, we get

Reorganize equations and get

Hence, the circuit realized the logarithm operation.