东南大学电子学院《电路基础》期末考试样卷及答案_东南大学电路基础期末

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Solve the following problems.(100 points)

1、(6 points)Find Uab in the circuit in Figure 1.3kΩI6mA2kΩ0.5I+-a+Uab-b Figure 120kΩ

2、(8 points)Find uo and io in the circuit in Figure 2.+-+io+uo-+1V80kΩ50kΩ-100kΩ10kΩFigure23、(6 points)In the circuit of Figure 3, readings of voltmeter V1，V2 and V3 are 10V, 18V and 6V, respectively.Please determinate the reading of the voltmeter ○V.○○○V1aVbLV2CRV3Figure34、(8 points)The resonant or tuner circuit of a radio is portrayed in Figure 4, where us1 represents a broadcast signal, given that R=10Ω，L=200μH, Us1rms=1.5mV，f1=1008kHz.If the circuit is resonant with signal us1, please determinate:(1)the value of C;(2)the quality factor Q of the circuit;(3)the current Irms;(4)the voltage Ucrms.iLR++-Cuc-us1Figure 4

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5、(8 points)A balanced three-phase circuit is shown in Figure 5.Calculate the phase currents and voltages in the delta-connected load Z, if UA2200V, Z=(24+j15)Ω，ZL=(1+j1)Ω.UA-UB+AZLA’Z-UC+BZLB’ZZ-+CZLC’Figure 5

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6、(8 points)Find the y parameters for the circuit in Figure 6.I15Ω-+I2+U1+10ΩU20.1U220I1-Figure 6-

7、(15 points)For the circuit in figure 7, find the rms value of the current i and the average power absorbed by the circuit, given that L10Ω, R15Ω,R220Ω,1/C20Ω,us(t)80100cost50cos(2t30)V.i+LR1usC-R2Figure 78、(12 points)In the circuit of Figure 8, determinate the value of Z that will absorb the maximum power and calculate the value of the maximum power.1ΩI12I1-+500Vrms+-4Ω250μFZω=103rad/sFigure 89、(14 points)If the switch in Figure 9 has been closed for a long time and is opened at t = 0, find uc, iL, uk for t ≥ 0.iL20Ω+1HS+20ΩuK20Ω-6A1F+uc-60V-Figure 910、(15 points)The switch in Figure 10 has been in position a for a long time.At t = 0, it moves to position b.Find i2(t)for t ≥ 0 and sketch it.(Attention: Please mark the key data in your graph.)9Ωab3Ωi12H2Ωi28H10Ω+60V2H-Figure 10

Solve the following problems.(100 points)

1、(6 points)Find Uab in the circuit in Figure 1.3kΩI6mA2kΩ0.5I+-a+Uab-b Figure 1Solution:

Uab= 9V2、(8 points)Find uo and io in the circuit in Figure 2.20kΩ+-+io+uo-+1V80kΩ50kΩ-100kΩ10kΩFigure 2Solution:

uo2.4V,io0.256mA3、(6 points)In the circuit of Figure 3, readings of voltmeter V1，V2 and V3 are 10V, 18V and 6V, respectively.Please determinate the reading of the voltmeter ○V.○○○V1aVbLV2CRV3Figure 3

Solution: The reading of the voltmeter ○V is 10V.4、(8 points)The resonant or tuner circuit of a radio is portrayed in Figure 4, where us1 represents a broadcast signal, given that R=10Ω，L=200μH, Us1rms=1.5mV，f1=1008kHz.If the circuit is resonant with signal us1, please determinate:(1)the value of C;(2)the quality factor Q of the circuit;(3)the current Irms;(4)the voltage Ucrms.iLR++-Cuc-us1Figure 4

Solution:

(1)C12.5pF

(2)Q12.6

(3)Irms150μA

(4)UCrms190 mV5、(8 points)A balanced three-phase circuit is shown in Figure 5.Calculate the phase currents and voltages in the delta-connected load Z, if UA2200V, Z=(24+j15)Ω，ZL=(1+j1)Ω.UA-UB+AZLA’Z-UC+BZLB’ZZ-+CZLC’Figure 5

Solution:

IA'B'11.74-3.69A,UA'B'33228.31V 

6、(8 points)Find the y parameters for the circuit in Figure 6.I15Ω-+I2+U1+10ΩU20.1U220I1-Figure 6-

Solution:

y0.2 0.02S 0.547、(15 points)For the circuit in figure 7, find the rms value of the current i and the average power absorbed by the circuit, given that L10Ω, R15Ω,R220Ω,1/C20Ω,us(t)80100cost50cos(2t30)V.i+LR1usC-R2Figure 7

Solution:

Irms22I0I12I23.224.71422.35726.17A

P=P0+P1+P2=256+333.28+50=639.28W

8、(12 points)In the circuit of Figure 8, determinate the value of Z that will absorb the maximum power and calculate the value of the maximum power.1ΩI12I1-+500Vrms+-4Ω250μFZω=103rad/sFigure 8 Solution:

If ZZeq5jΩ, Z will absorb the maximum average power.131UOC119.62249.7W 4Req4513The maximum average power is, Pmax9、(14 points)If the switch in Figure 9 has been closed for a long time and is opened at t = 0, find uc, iL, uk for t ≥ 0.iL20Ω+1HS+20ΩuK20Ω-6A1F+uc-60V-Figure 9

Solution:

iL1.5(01.5)e40t1.51.5e40tA, t0，uC120(60120)et/2012060et/20V, t0 uK20iLuC9030e40t60et/20V, t010、(15 points)The switch in Figure 10 has been in position a for a long time.At t = 0, it moves to position b.Find i2(t)for t ≥ 0 and sketch it.(Attention: Please mark the key data in your graph.)9Ωab3Ωi12H2Ωi28H10Ω+60V2H-Figure 10

Solution: i21.25et1.25e3tA

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