1. (30 points) A 8.00-mH inductor and a 2.00-ohm resistor are connected to a switch and a 5.00-V battery as shown in the figure. Initially the switch is closed on A and has been in that position for an infinite amount of time. (See Fig. 1)
(a) (2 points) In Fig. 1, what is the voltage difference Va - Vb across the inductor L?
(b) (2 points) In Fig. 1, what is the voltage difference Vb - Vc across the resistor R ?
(c )(5 points) In Fig. 1, what is the current i?
Suddenly at t = 0, the switch is thrown from A to B. (Fig. 2)
(d) (7) points) What is the value i of the current 450 m s after the switch is thrown to position B.(Fig. 2)
(e) (1 point) At the time you used in part (d), what is the voltage difference Vb - Vc across the resistor R ? (Fig. 2)
(f) (7 points) At the time you used in part (d), which is larger, the voltage Va at terminal a or the voltage Vb at terminal b? (Fig. 2)
(g) (6 points) At the time you used in part (d), what is the value of the voltage difference Va - Vb across the inductor? (Fig. 2)
| 1. (a) 0 volts since the sytem is in steady state (b) iR = 5 Volts (c) i = V/R = 2.5 A. (d) i = (5.00/R)e-0.00045/0.004 =2.23 A (e) iR = 4.46 V (f) b is higher (g) i = (5.00/R)e-0.00045/0.004 so that di/dt = -(1/0.004)(5.00/*R)e-0.0045/0.004 Thus, |Ldi/dt|= (0.008)(1/0.004)(5.00/R)e-0.00045/0.004 =4.46 V |
2. (34 points) In an RLC series circuit, the resistor, inductor and capacitor are connected in series with a voltage source given by:
,
where 70.0 V is the maximum voltage.
The current is measured to be:
1.60 A = maximum current. The value of the inductor is L = 210 mH and 
.
Please answer the following questions:
(a) (1 point) Is the phase angle positive or negative ? (Hint: Do not copy the result from the sample test.)
(b) (2 points) Does the current lead or lag the voltage of the source ?
(c) (6 points) What is the average power delivered to the circuit ?
(d) (5 points) How much average heat energy (in joules) is generated during a time period of 2 minutes ?
(e) (7 points) What is the resistance R of the circuit ?
(f) (2 points) Suppose the circuit is like your car radio. You can bring the circuit into resonance by changing value of the capacitor . Would you increase or decrease the capacitor value to bring this circuit into resonance ?
(g) (4 points) What is the value of the change 
in the capacitance to bring the circuit into resonance?
(h) (4 points) What is the value of the impedance Z at resonance?
(i) (3 points) What is the value of the average power at resonance ?
| Answers. (a) Negative (b) leads (c) Pav = ½ Im*Vmcos45=39.6 J/sec (d) energy = Power* time= 39.6*120 = 4725 J (e)R = Zcos45=(70/1.6)cos45 = 30 ohms (f) increase since Xc > XL (g) tan45 = (XL - XC )/R. Solve for C= 2.41x10-5 F. Then find Cf = 1/[(377)2L]= 3.35x10-5 F. The change in C =0.94x10-5 F (h) Z = R (i) ½ Im*Vmcos0 = 81 W, where Im = 70/R = 70/30 A= 2.33 A. Be careful!! |
3. (16 points)
A transverse wave on a string is described by :
(a) (4 points) Find the transverse velocity of a particle on the string at the time t = 0.20 sec and x =1.6 m.
(b) (2 points) Assume that up is in the positive y-direction and down is in the negative y-direction. Is the particle moving up or down at the value of t and x used to find the velocity in part (a)?
(c ) (4 points) Find the transverse acceleration of a particle on the string at the time t = 0.20 sec and x =1.6 m.
(d) (1 point) What is the wavelength ?
(e) (1 point) What is the period ?
(f) (1 point) What is the speed of the wave in the x-direction ?
(g) (3 points) Is the wave moving in the positive or negative x-direction?
| Answers. (a)  vT = 0.466 m/s (b) up (c) |
4. (8 points) Extra Credit
Two small speakers emit sound waves and are arranged as shown. Assume that speaker A has an output of 1.1 mW and speaker B has an output of 1.5 mW. Assume that both speakers emit sound.
(a)(4 points) How long (in seconds) does it take for the detector at point C to measure a total energy of 1.80x10 3 J ? Assume the detector has area AD= 1 cm2
(b)(4 points) Determine the sound intensity level (in dB) at point C.
| 4. (a) BC = 5 m and AC2=22
+ 42 so AC = 4.7 m
energy= (IA + IB)*area
*time where area = 10-4 m2 |
