45. The peak value of an alternating current in a 1500W device is 5.4 A. What is the rms voltage across?


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1 PHYS Practice Problems hapters 8 hapter The peak value of an alternating current in a 5W device is 5.4 A. What is the rms voltage across? The power and current can be used to find the peak voltage, and then the rms voltage can be found from the peak voltage. hapter 9. peak I peak P 5 W P = I rmsvrms = Vrms Vrms = = = 3.9 V I 5.4 A 5. Eight 7.W hristmas tree lights are connected in series to each other and to a V source. What is the resistance of each bulb? Vtot Each bulb will get oneeighth of the total voltage, and so V =. Use that voltage and the bulb 8 power dissipated by each bulb to calculate the resistance of a bulb. P bulb ( V) V V V bulb bulb tot = R = = = = 7Ω R P 64P W. What is the net resistance of the circuit connected to the battery in Fig. 94? Each resistance has R =.8 kω. R The resistors have been numbered in the accompanying diagram to help in the analysis. R and R are in series with an equivalent resistance of R = R + R = R. This combination is in parallel with R 3, with an equivalent resistance of R = R 3 + =. This combination is in 3 R R 5 series with R, with an equivalent resistance of R = R + R = R. This combination is in parallel with R, with an equivalent resistance of R = R = 8 R 5R final equivalent resistance.. Finally, this combination is in series with R, and we calculate the 6 R = R + R = R = Ω = Ω k 4.55 k eq R R 5 R3 R4 R 6
2 7. Determine the magnitude and directions of the currents through R and R in Fig Because there are no resistors in the bottom branch, it is possible to write Kirchhoff loop equations that only have one current term, making them easier to solve. To find the current through R, go around the outer loop counterclockwise, starting at the lower left corner. V3 + V 6.V + 9.V V I R + V = I = = =.68A, left 3 R Ω To find the current through R, go around the lower loop counterclockwise, starting at the lower left corner. V3 6.V V I R = I = = =.4A, left 3 R 5 Ω 5. In Fig (same as Fig. 9 a), the total resistance is 5. kω, and the battery s emf is 4. V. If the time constant is measured to be 35. µs, calculate (a) the total capacitance of the circuit and (b) the time it takes for the voltage across the resistor to reach 6. V after the switch is closed. (a) From Eq. 97, the product R is equal to the time constant s τ 9 τ = R = = =.33 F R 3 5. Ω (b) Since the battery has an EMF of 4. V, if the voltage across the resistor is 6. V, the voltage across the capacitor will be 8. V as it charges. Use the expression for the voltage across a charging capacitor. hapter. τ V e e V t V E τ E t / t / τ = E = = ln V 8. V t = τ = ( ) = E 4. V 6 5 ln 35. s ln.4 s 5. The force on a wire carrying 8.75 A is a maximum of.8 N when placed between the pole faces of a magnet. If the pole faces are 55.5 cm in diameter, what is the approximate strength of the magnetic field? Use Eq. . The length of wire in the Bfield is the same as the diameter of the pole faces. F IlB B F.8 N max = = = = max Il ( 8.75A)(.555m).64 T
3 . Determine the direction of B r for each case in Fig. 5, where F r represents the maximum magnetic force on a positively charged particle moving with velocity r v. The right hand rule applied to the velocity and magnetic field would give the direction of the force. Use this to determine the direction of the magnetic field given the velocity and the force. (a) downward (b) inward into the paper (c) right 5. An electron experiences the greatest force as it travels.9 x 6 m/s in a magnetic field when it is moving northward. The force is upward and of magnitude 7. x 3 N What are the magnitude and direction of the magnetic field? The magnetic field can be found from Eq. 4, and the direction is found from the right hand rule. Remember that the charge is negative. 3 Fmax 7. N Fmax = qvb B = = =.55T 9 6 qv (.6 )(.9 m s) The direction would have to be East for the right hand rule, applied to the velocity and the magnetic field, to give the proper direction of force. 3. Determine the magnitude and direction of the force between two parallel wires 35 m long and 6. cm apart, each carrying 5 A in the same direction. Since the currents are parallel, the force on each wire will be attractive, toward the other wire. Use Eq. 7 to calculate the magnitude of the force. 7 ( 4π T m A) ( 5A) ( 35m ) π µ I I F = l = = π d 6. m.79 N, attractive 36. A straight stream of protons passes a given point in space at a rate of.5 x 9 protons/s. What magnetic field do they produce. m from the beam? The stream of protons constitutes a current, whose magnitude is found by multiplying the proton rate times the charge of a proton. Then use Eq. 6 to calculate the magnetic field. B µ I ( 4π T m A)(.5 protons s)(.6 proton ) 7 stream = = =.4 T π r π. m
4 hapter. 5. A.cm diameter loop of wire is initially oriented perpendicular to a.5t magnetic field. The loop is rotated so that its plane is parallel to the field direction in. s. What is the average induced emf in the loop? The flux changes because the loop rotates. The angle changes from o to 9 o. The magnitude of the average induced emf is given by Eq. a. Φ AB cosθ π B E = = = t t (.6 m).5 T ( ) π = =.s o o (.6 m).5t ( cos 9 cos ) 84.8 mv.s 7. In Fig. , the rod moves with a speed of.6 m/s, is 3. cm long, and has a resistance of.5 Ω. The magnetic field is.35 T, and the resistance of the Ushaped conductor is 5. Ω at a given instant. alculate (a) the induced emf, (b) the current in the Ushaped conductor, and (c) the external force needed to keep the rod s velocity constant at that instant. (a) (b) (c) Because the velocity is perpendicular to the magnetic field and the rod, we find the induced emf from Eq. 3. E = Blv =.35T.3m.6m s =.68V Find the induced current from Ohm s law. E.68 V 6. 3 I = = = A R 7.5Ω The induced current in the rod will be down. Because this current is in an upward magnetic field, there will be a magnetic force to the left. To keep the rod moving, there must be an equal external force to the right, given by Eq. . F = IlB = = A.3 m.35 T 6.4 N. A simple generator is used to generate a peak output voltage of 4. V. The square armature consists of windings that are 6. cm on a side and rotates in a field of.4 T at a rate of 6. rev/s. How many loops of wire should be wound on the square armature? We find the number of turns from Eq. 5. The factor multiplying the sine term is the peak output voltage. E E 4. V = = = = 4. 4 loops peak NBω A N peak B ω A (.4T) π rad rev ( 6 rev s)(.6 m)
5 3. A stepup transformer increases 5 V to V. What is the current in the secondary coil as compared to the primary coil? Use Eqs. 6 and 7 to relate the voltage and current ratios. V N I S S S N V P S I I P S V 5V P = ; = = = = =.8 V N I N V I I V V P P P S P S P S 4. How many turns of wire would be required to make a 3mH inductance out of a 3. cmlong airfilled coil with a diameter of 5. cm? Use the relationship for the inductance of a solenoid, as given in Example 4. µ N A Ll L = N = = l (.3 H)(.3 m) 7 ( 4 T m A) (.6 m) µ A π π = 383 turns 55. At what frequency will a.4µf capacitor have a reactance of 6.7 kω? We find the frequency from Eq. b for the reactance of a capacitor. X = f = = = π f π X π F 3 6 ( Ω )( ) 9.9 Hz 66. A.5kΩ resistor in series with a 4mH inductor is driven by an ac power supply. At what frequency is the impedance double that of the impedance of 6 Hz? The impedance is given by Eq. 5 with no capacitive reactance. L ( π ) Z = R + X = R + fl. 6 Z = Z R + 4π f L = R + 4π 6 Hz L f R + 4π f L = 4 R + 4π 6 Hz L = 4R + 6π 6 Hz L =.64 khz + π Ω 3R 6 6 Hz L 3R 3 5 f = = + 4( 6 Hz) = Hz L L 4π 4π 4π.4 H
6 hapter. 4. In an EM wave traveling west, the B field oscillates vertically and has a frequency of 8. khz and an rms strength of 6.75 x 9 T. What are the frequency and rms strength of the electric field, and what is its direction? (See Fig. 7) The frequency of the two fields must be the same: 8. khz. The rms strength of the electric field is 8 9 Erms = cbrms = 3. m s 6.75 T =.3V m. The electric field is perpendicular to both the direction of travel and the magnetic field, so the electric field oscillates along the horizontal northsouth line. 8. The E r field in an EM wave in free space has a peak of.8 mv/m. What is the average rate at which this wave carries energy across unit area per unit time? The energy per unit area per unit time is S = cε E = = m s 8.85 N m.8 V m 6.3 W m. 3. The oscillator of a 96.MHz FM station has an inductance of.8 µh. What value must the capacitance be? We find the capacitance from the resonant frequency: f = π L ; Hz =, which gives.53 F.53pF. 6 = = π (.8 H)
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