《电路基础》样卷+答案

1、( 6 points) Find Uab in the circuit in Figure 1.

3kΩI6mA2kΩ0.5I+-a+Uab-b

Figure 1Solution:

Uab= 9V

2、( 8 points) Find uo and io in the circuit in Figure 2.

20kΩ+-+io+uo-

+1V80kΩ50kΩ-100kΩ10kΩFigure 2Solution:

uo2.4V, io0.256mA

3、( 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126.6 (3)Irms150μA (4)UCrms190 mV

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

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

4 0.5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 7

Solution:

22IrmsI0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Zeq5jΩ, Z will absorb the maximum average power. 131UOC119.62249.7W 4Req4513The maximum average power is, Pmax

9、( 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0

10、( 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