Download Giambattista, Ch. 18 Problems: 1, 12, 22, 23, 40, 42, 44, 52

Giambattista, Ch. 18 Problems: 1, 12, 22, 23, 40, 42, 44, 52, 55, 57, 60, 61 Note that in the printing of some of these equations, some symbols don’t always come through. “delta V” sometimes looks like a % sign. 1.Strategy Use the definition of electric current. (Current is defined as “charge per time”.)
Solution Compute the total charge.
12. (a) Strategy Calculate the work done using Eq. (18-2) and the definition of electric current.
Alternatively, you could think about Part b first, and multiply power by time to get energy.
Solution Compute the amount of electrical energy supplied by the solar cell.
(b) Strategy The power is equal to the rate at which the solar cell supplies electrical energy.
Solution Find the average power by dividing the energy supplied by the time.
22. (a) Strategy Use the definition of resistance.
Solution Compute the resistance (i.e., Ohm’s Law)
(b) Strategy and Solution The current flows from right to left through the battery (from low to
high potential). Thus, the current flows from left to right through the resistor.
23. Strategy Use Eq. (18-8). The cross-sectional areas of the wires are given by
Solution Form a proportion to find the ratio of diameters.
40. (a) Strategy Use Eqs. (18-13) and (18-17). You ought to redraw the circuit so that you can see
what things are in parallel or series, one pair at a time.
Solution Compute the resistance between points A and B.
(b) Strategy Label the currents on a diagram. Use Kirchhoff’s rules.
I
Solution The current through the emf is
where the currents labeled 1
I1
and 2 are shown in the diagram. Applying the loop
rule, we have
Solve
for the current through the
resistor,
I2
R2 = 24 Ω
R1 = 12 Ω
I
42. Strategy Use Eqs. (18-13) and (18-17). The current through the
through the emf,
resistor is the same as that
Solution Find
Find R.
44. (a) Strategy
Use Eqs. (18-15) and (18-18). Capacitors “add up” in the opposite
way that resistors do, so that when they’re in series they increase the total capacitance, and
when they’re in parallel they decrease the total capacitance.
Solution Find the unknown capacitance.
(b) Strategy The charge on the
Use Eq.
(17-14).
capacitor is the same charge as on the equivalent capacitor.
Solution Find the charge.
52. Strategy Use Kirchhoff’s rules. Let
be the top branch,
be the middle branch, and
be the
bottom branch. Assume that each current flows right to left.
NOTE: This strategy and solution is just one way of setting up the problem. You could label
the currents differently and even have them go different directions, but the equations you
create from these will all solve to the same result. The algebra is messy, but not technically
difficult.
Solution Find the current in each branch of the circuit.
(1)
(2)
Substitute (1) into (2).
Solve (3) for
Solve for
Calculate
Calculate
and substitute into (4).
(3)
To two significant figures, the currents are:
Branch
I (mA)
Direction
Top
85
right to left
Middle
56
left to right
Bottom
29
left to right
55. Strategy Use Kirchhoff’s rules. Let the current on the left be I, the one in the middle be
one on the right be
and the
flows downward.
NOTE: Same as above; the details of what you call each current and which directions you use
are up to you. You just need to make sure that you’re consistent. Also note that in the
printing of some of these equations, some symbols don’t always come through. “delta V”
sometimes looks like a % sign.
Solution Find the unknown emf and the unknown resistor.
Loop ABCFA:
Loop ABCDEFA:
57. Strategy Use Eq. (18-20).
Solution Compute the power (P=IV) dissipated by the resistor.
60. Strategy Use Eq. (18-21b).
Solution Compute the resistance of the lightbulb.
61. Strategy and Solution Yes; the power rating can be determined by