8. A flashlight bulb works as follows. It is simply a resistive wire connect to a source of emf. The source drives a current through the wire which, through Joule heating, gets hot and then radiates energy. A schematic picture is shown below:

13. There are three bulbs. The top 2 bulbs are in series and are in parallel with the lower bulb. Since all three bulbs are claimed to be identical, each has the same resistance. Therefore,

the current through the top branch must be 1/2 that of the current through the bottom branch.

15. To do this problem, we could, by brute force, simplify the circuit, calculate resistances, and determine currents. Or, we can use words. Let's use words.

• Start at the battery. The first resistor, R(1), sees the entire current I.
• When we reach resistors, R(2), R(3), and R(4), we note that they are in parallel and so the current must split. Since all resistors have the same resistance, the current splits into 3 equal parts, i.e., each of 2, 3, and 4, sees I/3.
• When we reach R(5), there is only path the current can follow and so, R(5) sees all of the current, I.
• R(6) sees no current, because it is not part of a closed circuit.
• When we reach R(7), we see that it is in parallel with a branch which is 0 resistance. The current splits and takes the "easier" path (to the ground). R(7) does not see any current because its ends are at the same voltage.

Problems

1. Before the wire is clipped, the circuit contains a battery of voltage 12 Volts and 2 resistors (light bulbs) or resistance 8 ohms and 12 ohms, respectively. The light bulbs are in series.

The resultant current is then, I = V/R = 12 Volts/(8 ohms + 12 ohms) = 0.6 A

I = V/R = 12 Volts/8 ohms = 1.5 A

3. The 4 20 ohm lamps are in series and so an equivalent resistor has a resistance of

R = 20 ohm + 20 ohm + 20 ohm + 20 ohm = 80 ohm

Since P = IV, when the current doubles, the power delivered to the circuit by the battery doubles.

7. An indiviual lamp operating at 100 Volts produces 100 W of power. This means that it consumes I = P/V = 100 W/100 Volts = 1 A when producing 100 W of power.

So, for all 5 lamps to produce 100 W of power means that the battery must supply 5 x 1 A = 5 A.

11. If you start at point A and follow the wire until you reach point B, you notice that you have two possible paths. The two paths are in fact parallel. The paths are

• Point A ==> 2 ohm --> 3 ohm (top one) --> 1 ohm ==> point B
• Point A ==> 3 ohm (bottom one) ==> Point B

The total resistance ofthe circuit is then

1/R(total) = 1/(6 ohm) + 1/(3 ohm) = 1/(2 ohm) ===> R(total) = 2 ohm

14,15,16.

• A and B: (1) see the 10 ohm and 10 ohm in parallel ---> 1/R = 1/(10 ohm) + 1/(10 ohm) --> 1/R = 1/(5 ohm) ---> R = 5 ohm. (2) These combine with the 2 ohm resistor in series ===> total resistance = 7 ohm.
• A and C: (1) only see the 10 ohm and 10 ohm resistors in parallel ===> total resistance = 5 ohm. The 2 ohm resistor is not in the circuit.
• B and C: The 2 10 ohm resistors are in parallel with a zero resistance wire ===> the current will by-pass them and the circuit only "sees' the 2 ohm resistor

20. The idea is to determine the total voltage drop through the circuit. This must be the voltage supplied by the battery. To accomplish, we need to determine the current which flows in the circuit, because we want to find the voltage drop at each resistor from V = IR.

• The Ammeter measures the current through the 12 ohm resistor. Therefore, the voltage drop through the 12 ohm resistor is V = I R = 1 A x 12 ohm = 12 Volts. The current through the 6 ohm resistor, is then I = V/R = 12 Volts/6 ohm = 2 A. The potential drop is the same as for the 12 ohm resistor because they are in parallel. he total current through the parallel resistors is 1 A + 2 A = 3 A. This then must be the current through the other resistors.
• The battery's voltage must be:
  V = 3 A x 4 ohm + 12 Volts + 3 A x 2 ohm = 12 Volts + 12 Volts + 6 Volts = 30 Volts.

23. The circuit contains a battery which supplies 60 Volts, a lamp which uses 20 Volts when producing 80 W or power and a resistor R in series. The question is what must R be in order for the lamp to run properly? What is meant by "running properly"? Well, we want the current amount of current to flow through the lamp. What is this I? For recommended operation of the lamp I = P/V = 80 W/20 Volts = 4 A. So, for this current, we want the voltage drop across the lamp plus the voltage drop across the resistor to be the same as the emf of the battery:

60 Volts = 20 Volts + 4 A x R ===> R = 40 Volts/4 A = 10 ohm

31. You need to simplify the circuit in the following steps:

• the 9 ohm and 3 ohm resistors are in parallel and combine to 1/R = 1/(9 ohm) + 1/(3 ohm) = 4/(9 0hm) ===> R = 9/4 ohm
  • his combination is in parallel with the 18 ohm resistor and so the equivalent resistance is 1/R = 1/(18 ohm) + 1/(9/4 ohm) = 1/(2 ohm) ==> 2 ohm
    • This combination is in parallel with the 3 ohm and 6 ohm resistors for an equivalent resistance of 1/R = 1/(2 ohm) + 1/(6 ohm) + 1/(3 ohm) = 1/(1 ohm) ===> R = 1 ohm

This 1 ohm resistor is in series with the 4/5 ohm resistor and so the total resistance is 1.8 ohm from A to B.

37.

38. To begin, figure out the equivalent resistance of the circuit.

• R = 5.5 ohm + "stuff" + 0.5 ohm

  What is the "stuff"? It is the 4 ohm, 9 ohm, 6 ohm, and 4 ohm resistors. We see that the resistors are bascially in two parallel branches: (1) the lower 4 ohm resistor; (2) the combination of 4 ohm and 9 ohm resistors (which are in parallel with each other) in series with the 6 ohm resistor. The 4 ohm and 9 ohm resistors have an equivalent resistance of 1/R = 1/(4 ohm) + 1/(9 ohm) = 36/13 ohm = 2.8 ohm and so the top branch of the stuff has an equivalent resistance of R = 6 ohm + 2.8 ohm = 8.8 ohm. The complete parallel configuration has an equivalent resistance of

  1/R = 1/(4 ohm) + 1/(8.8 ohm) = 3.2/(8.8 ohm) ==> R = 8.8/3.2 ohm = 2.8 ohm

  The total resistance of the circuit is then R = 5.5 ohm + 2.8 ohm + 0.5 ohm = 8.8 ohm.

  a. The current in the circuit is then I = V/R = 9 Volts/8.8 ohm = 1.0 A.

  b. The power supplied to the circuit is P = IV = 1.0 A x 9.0 V = 9 W.

  c. The terminal voltage is V = 9 Volts - 1.0 A x 0.50 ohm = 8.5 Volts

  Note -- From hereon I simply scanned in hand-written solutions. Downloading the following pages might be slow.

50.

51a, 51b, 51c, 51d

52.