Phys 4 b exam 2 ch. 25 problems

A ductile metal wire has resistance R. What will be the resistance of this wire in terms of R if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)

both L and A changes L changes to 3L A changes to A/3 (because A = V/L so the new A = V/3L) The equation becomes R = p(3L)/ (A/3) R=9pL/A so 9R

A tightly coiled spring having 75 coils, each 3.50 cm in diameter, is made of insulated metal wire 3.25 mm in diameter. An ohmmeter connected across its opposite ends reads 1.74 ohms. What is the resistivity of the metal?

1.74 = p(L/A) L = 75π(.035) L=8.25 m A= (π/4)d² = 8.3 x 10⁻⁶ p=1.74A/L p=1.75x10⁻⁶

A 25.0 ohm bulb is connected across the terminals of a 12.0-V battery having of internal resistance of 3.5 ohms. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?

P = I²R I²r / I²(r+R) = r / r+R = 3.5 / (25+3.5) = .123 number 49

A cylindrical copper cable 1.50 km long is connected across a 220.0-V potential difference. (a) What should be its diameter so that it produces heat at a rate of 75.0 W? (b) What is the electric field inside the cable under these conditions?

P = V²/R = V²/(pL/A) = V²A/pL= P A=PpL/V² πr² = PpL/V² r= √(Ppl/πV²) r=9.21x10⁻⁵ d=2r=.184mm b) E=V/L E = (220 V)/(1500 m) =0.147 V/m #74

The circuit shown in Fig. E25.32 contains two batteries, each with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit (magnitude and direction);

Potential decreases by IR when going through a resistor in the direction of the current and increases byε when passing through an emf in the direction from the − to + terminal. a) 16V -8V -(3+5 +9)I =0 I = .47 b) V(ab) = 16V -(1.6)(.47) = 15 2 V c) Vab - IR = 15.2 - (.47)(9) number 32

(a) At room temperature what is the strength of the electric field in a 12-gauge copper wire (diameter 2.05 mm) that is needed to cause a 2.75-A current to flow? (b) What field would be needed if the wire were made of silver instead?

a) E = pJ =p(i/A) =p(2.75)/(.00205²π/4) =1.43x10²² b) E(s) = E(c) (p(s)/p(c))

A lightning bolt strikes one end of a steel lightning rod, producing a 15,000-A current burst that lasts for 65 µs. The rod is 2.0 m long and 1.8 cm in diameter, and its other end is connected to the ground by 35 m of 8.0-mm-diameter copper wire. (a) Find the potential difference between the top of the steel rod and the lower end of the copper wire during the current burst. (b) Find the total energy deposited in the rod and wire by the current burst.

a) V = IR = I(Rsteel + Rcopper) =(15000)(1.57x10⁻³+.012) = 204 V b) E = Pt =i²Rt = (15000)²(1.57x10⁻³+.012) (65x10⁻⁶) = 199J

Consider the circuit shown in Fig. E25.28. The terminal voltage of the 24.0-V battery is 21.2 V. What are (a) the internal resistance r of the battery and (b) the resistance R of the circuit resistor?

a) V(ab) = ε - Ir 21.2 = 24 -Ir r=.7 b) V=IR 21.2 = 4R R=5.3 number 28

What is the potential difference V(ad) in the circuit of Fig. P25.68? (b) What is the terminal voltage of the 4.00-V battery? (c) A battery with emf 10.30 V and internal resistance .5 ohms is inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00-V battery. What is the difference of potential V(bc) between the terminals of the 4.00-V battery now?

can do either: ε₁-.5i-8i-6i-ε₂-.5i-9i=0 or I=(8-4)/24 = .167 a) V(ad) = Va - Vd Vd + 8 - 8.5i = Va Va-Vd = 8-8.5(.167) Vad=6.58 b) Vbc = Vb-Vc Vb - 4V -.167 (.5) = Vc Vb-Vc= 4 +.167(.5) Vbc = 4.0835 c) direction switches with new batter recalculate I and R new answer is 3.87

A copper wire has a square cross section 2.3 mm on a side. The wire is 4.0 m long and carries a current of 3.6 A. The density of free electrons is 8.5x10²⁸ Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

cross sect area = 5.29x10⁻⁶ m² a) J=i/A =6.8 x 10⁵ b) E=pJ =(1.72x10⁸)(6.8x10⁵) = .012 /m c) x/t = v so t = x/v v=J/nq t= x / (J/nq) t = L n q A / I =8x10⁴ s

In the circuit of Fig. P25.79, find (a) the current through the 8.0-ohm resistor and (b) the total rate of dissipation of electrical energy in the 8.0-ohm resistor and in the internal resistance of the batteries. (c) In one of the batteries, chemical energy is being converted into electrical energy. In which one is this happening, and at what rate? (d) In one of the batteries, electrical energy is being converted into chemical energy. In which one is this happening, and at what rate? (e) Show that the overall rate of production of electrical energy equals the overall rate of consumption of electrical energy in the circuit.

a) 12 - i - 8i - 8 - i = 0 12-8-10i=0 i=.4A b) P=i²R P=10(.4) P=1.6 W c) Chemical energy is converted to electrical energy in a battery when the current goes through the battery from the negative to the positive terminal, so the electrical energy of the charges increases as the current passes through. This happens in the 12.0-V battery, and the rate of production of electrical energy is P=εi = (12V)(.40 A) = 4.8 W (d) Electrical energy is converted to chemical energy in a battery when the current goes through the battery from the positive to the negative terminal, so the electrical energy of the charges decreases as the current passes through. This happens in the 8.0-V battery, and the rate of consumption of electrical energy is 3.2 W

A current-carrying gold wire has diameter 0.84 mm. The electric field in the wire is .49 V/m What are (a) the current carried by the wire; (b) the potential difference between two points in the wire 6.4 m apart; (c) the resistance of a 6.4-m length of this wire?

a) E = V/L = p(i/A) E=p(i/A) I=E(A)/p i=11.1 b) V=Ed V=3.14 c) V=IR R=.28 #23

A cylindrical tungsten filament 15.0 cm long with a diameter of 1.00 mm is to be used in a machine for which the temperature will range from room temperature 20°C up to 120°C. It will carry a current of 12.5 A at all temperatures (consult Tables 25.1 and 25.2). (a) What will be the maximum electric field in this filament, and (b) what will be its resistance with that field? (c) What will be the maximum potential drop over the full length of the filament?

a) E max is at higher temperature p(120) = p(20)(1+α(T-T₀)) p(120) = 7.61x10⁻⁸ E = pJ = p(i/A) =1.21 b) R = pL/A = pL/πr² R=(7.61x10⁸)(.15)/π(.0005)² R= .0145 c) V=Ed V=(1.21)(.15) V= .1815 V

The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours A battery can supply a current of 50 A for 1.0 h, or 25 A for 2.0 h, and so on. (a) What total energy can be supplied by a 12-V, 60-A # h battery if its internal resistance is negligible? (b) What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (See Section 17.6; the density of gasoline is 900kg/m³ ) (c) If a generator with an average electrical power output of 0.45 kW is connected to the battery, how much time will be required for it to charge the battery fully?

a) Energy = Power x time VI = 12 x 60 = 720 720 x 3600 s = 2.6 x 10⁶ b)V = m / p = .0565 / 900 = 6.3 x 10⁵ (1000L/m³) c) U = Pt t = U / P =2.6 x 10⁶ J / 450 W = 5800 s = 97 min number 47

An idealized voltmeter is connected across the terminals of a 15.0-V battery, and a 75.0 ohm appliance is also connected across its terminals. If the voltmeter reads 11.3 V: (a) how much power is being dissipated by the appliance, and (b) what is the internal resistance of the battery?

a) I = V/R = 11.3 / 75 = .151 P = V I = 1.7 W b) ε - Ir = Vab 15 - .151r = 11.3

A hollow aluminum cylinder is 2.50 m long and has an inner radius of 3.20 cm and an outer radius of 4.60 cm. Treat each surface (inner, outer, and the two end faces) as an equipotential surface. At room temperature, what will an ohmmeter read if it is connected between (a) the opposite faces and (b) the inner and outer surfaces?

a) R = pL/A =pL/π(b²-a²) =pL/π(.0460²-.0320²) = 2x10⁻⁵ b) dR=pdr/2πrL p/2πL integral of 1/rdr from a b R = (p/2πL) ln(a/b) = 6.35x10⁻¹⁰ ohms #24

An electrical conductor designed to carry large currents has a circular cross section 2.50 mm in diameter and is 14.0 m long. The resistance between its ends is .104 ohms. (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is 1.28 V /m, what is the total current? (c) If the material has 8.5 x 10²³ free electrons per cubic meter, find the average drift speed under the conditions of part (b).

cross sectional area = (π/4)d²= 4.9x 10⁻⁶) a) R=p(L/A) .104 = p (14/4.9 x 10⁻⁶) p = 3.64 x 10⁻⁸ b) V = Ed = 1.28 x 14 = 17.9 V V=IR 17.9 = .104I ---> I = 172A c) J= i / A = 172 / (4.9x 10⁻⁶) J = 2.58 x 10⁻³ J = nqV 2.58 x 10⁻³ = (8.5x10²⁸)(1.6x10⁻¹⁹)V V = 2.58 x 10⁻³

A plastic tube 25.0 m long and 3.00 cm in diameter is dipped into a silver solution, depositing a layer of silver 0.100 mm thick uniformly over the outer surface of the tube. If this coated tube is then connected across a 12.0-V battery, what will be the current?

cross sectional area = 2πrT =(.03)π x (.0001) = 9.4x10⁻⁶ V=IR -> I=V/R =V/(pL/A) = (V)(A) /(pL) =307 A false solution says 410 problem 56

In the circuit analyzed in Example 25.8 the 4.0-ohm resistor is replaced by a 8.0- resistor, as in Example 25.9. (a) Calculate the rate of conversion of chemical energy to electrical energy in the battery. How does your answer compare to the result calculated in Example 25.8? (b) Calculate the rate of electrical energy dissipation in the internal resistance of the battery. How does your answer compare to the result calculated in Example 25.8? (c) Use the results of parts (a) and (b) to calculate the net power output of the battery. How does your result compare to the electrical power dissipated in the 8.0-ohm resistor as calculated for this circuit in Example 25.9?

first find current i = ε / r+R = 1.2 A then find power P=εi = 12 v x 1.2 = 14.4 W less b) P=I²r =1.2² x 2 = 2.9 W which is less than the amount found in 25.8 c) 14.4 - 2.9 = 11.5 W

In the circuit in Fig. E25.51, find (a) the rate of conversion of internal (chemical) energy to electrical energy within the battery; (b) the rate of dissipation of electrical energy in the battery; (c) the rate of dissipation of electrical energy in the external resistor.

first use circuit to calculate I ε - Ir - IR = 0 12 = I(r+R) I = 2 a) P = εI = 12 V x 2 A = 24 W b) P= I²r = (2.0 A)²(1) = 4 W c) P= I²R = (2²)(5) = 20 W

Consider a resistor with length L, uniform cross-sectional area A, and uniform resistivity that is carrying a current with uniform current density J. Use Eq. (25.18) to find the electrical power dissipated per unit volume, p. Express your result in terms of (a) E and J; (b) J and p; (c) E and p

for all parts of the problem use p = P / LA and solve for p (a) E is related to V and J is related to I, so use P=VI p=P/LA = E L J A / L A = EJ b) J and P --> p = P / LA = I²R / LA number 43

Consider the circuit of Fig. E25.32. (a) What is the total rate at which electrical energy is dissipated in the 5.0-ohm and 9.0-ohm resistors? (b) What is the power output of the 16.0-V battery? (c) At what rate is electrical energy being converted to other forms in the 8.0-V battery? (d) Show that the power output of the 16.0-V battery equals the overall rate of dissipation of electrical energy in the rest of the circuit.

i = 16-8/17 = .47 a) P=I²R = (.47)²(1.1)=1.1 W P of 9ohm resistor is 3 w b) P(₁₆) = εi-i²r =16(.47) - (.47)²(1.6) 7.2 W c) P(₈) = εi + i²r =8(.47) + (.47)²(1.6) 4.1 W d) b = a + c

An idealized ammeter is connected to a battery as shown in Fig. E25.30. Find (a) the reading of the ammeter, (b) the current through the 4.00-ohm resistor, (c) the terminal voltage of the battery

remember theres no potential drop across idealized ammeter and it acts like a short circuit across terminals of battery (removes other resistor) a) ε-i(2) = 0 10=2i i=5 amps b) no current goes through the resistor as it all goes to the ammeter so V=0 c) 0 because theres no potential drop across the ammeter

The potential difference across the terminals of a battery is 8.40 V when there is a current of 1.50 A in the battery from the negative to the positive terminal. When the current is 3.50 A in the reverse direction, the potential difference becomes 10.20 V. (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

set up two equations ε - 1.5r=8.4 ε - 3.5r = 10.4 solve for r and then ε solution manuals wrong problem 69