21. Two resistors are connected in parallel as shown.
a) What is the current through each?
b) What total current does the battery supply?
c) What is the effective resistance in the circuit?

a) i1 = E/R i1 = (24)/(8) i1 = 3.0 A i2 = E/R i2 = (24)/(18) i2 = 1.33 A

b) i = i1 + i2 i = 3.0 + 1.33 i = 4.33 A

c) R = E/i R = (24)/(4).33 R = 5.54 Ô

Now THAT is a real shocker! We have taken not one, but TWO resistors, hooked them together, and found that the total resistance is less than either one individually. Is there an error in reasoning?

Of course not. Your favorite physics teacher is looking out for you. Just for a moment, instead of thinking of resistors as impediments to the flow of electricity, think of them as pathways. Between the top and bottom wire of problem 14 there will be no flow of electrons until we begin to put pathways in place. The more pathways, the more current and the less the resistance.

Think of resistors as check-out registers at a supermarket. When only one register is open, the line crawls. As new check-out people are brought in, the flow of groceries increases, no matter how slow (how large a resistance) a particular checker is.

22. What is the current through each resistor? What is the total current? What is the effective resistance?

a) i1 = E/R1 i2 = E/R2

b) i = i1 + i2 i = E/R1 + E/R2 i = E[(1/R1) + (1/R2)]

c) R = E/i R = E/[ E[(1/R1) + (1/R2)] ] 1/R = 1/R1 + 1/R2

We now have two working equations:

When resistors are in series: R = R₁ + R₂
When in parallel: 1/R = 1/R₁ + 1/R₂

Note that this is the reverse of the situation with capacitors, yet it is just as logical.

23. Find the equivalent resistances:

a) 1/R = (1)/(20) + (1)/(10) R = 6.67 Ô

b) R = R1 + R2 R = 16 + 30 R = 46 Ô

c) 1/R = (1)/(15) + 1/(8 + 12) R = 8.57 Ô

24. Again, find the equivalent resistances:

a) R1 = 20 + 11 R1 = 31 Ô 1/R2 = (1)/(31) + (1)/(40) R2 = 17.5 Ô RT = 18 + 17.5 RT = 35.5 Ô

b) 1/R1 = (1)/(19) + (1)/(25) R1 = 10.8 R2 = 10.8 + 15 + 7 R2 = 32.8 1/R = (1)/(50) + (1)/(32).8 R = 19.8 Ô

25. The current through the 12 Ô is 0.3 A. What is the current through the 4 Ô? What is the potential of the battery?

a) E = E8, 12 E8, 12 = iR E8, 12 = (.3)(8 + 12) E8, 12 = 6.0 V

b) E4 Ô = E E4 Ô = 6.0 V i4 = E4/R4 i4 = (6.0)/(4) i4 = 1.5 A

26. The current through the 160 Ô resistor is 0.3 A.
a) What is the potential across the 100 Ô?
b) What is the potential of the battery?

a) Total current through the 50 Ô - 100 Ô branch is also 0.3 A. Hence: E50 - 100 = iR E50 - 100 = (.3)[ 1/[(1/50) + (1/100)] ] E50 - 100 = 10.0 V

b) Ebattery = E160 Ô + E50 - 100 Ô Ebattery = (.3)(160) + 10 Ebattery = 58.0 V

27. The current in the 8 Ô is 2 A. What is the current in the 4 Ô?

E = iR E = 2[ 1/[(1/4) + (1/12)] ] E = 6.0 V i = E/R i = (6.0)/(4) i = 1.5 A note: 6.0 V is also the E across the 4 Ô

28. What is the potential across the 5 K resistor?

i = E/R i = 6/[(5 + 10) * 103] i = 4.0 * 10-4 A E = iR E = (4 * 10-4)(5 * 103) E = 2.0 V

29. What is the potential across the 400 Ô?

RT = 200 + 1/[(1/200) + (1/400)] RT = 333 Ô iT = E/R iT = (60)/(333) iT = .180 E200 - 400 = iR E200 - 400 = .180[ 1/[(1/200) + (1/400)] ] E200 - 400 = 24.0 V

30. What is the potential across R₁?

i = ET/RT i = E/(R1 + R2) E1 = iR1 E1 = [E/(R1 + R2)]R1 E1 = E[(R1)/(R1 + R2)] note: In the series circuit, each resistor has voltage proportional to its resistance. This is called a "voltage divider."

31. The potential across the 18 Ô is 6 V. What is it across the whole circuit?

Using the relationship from #30 E1/ET = R1/RT 6/ET = 18/(18 + 50 + 35) ET = 34.3 V

32. 6.0 amps flows into a branched circuit as shown. What current will go through the 8 ohm resistor?

Two methods of solution: method 1: ET = iRT ET = (6)[ 1/[(1/8) + (1/16)] ] ET = 32 V i8 = ET/R i8 = (32)/(8) i8 = 4.0 A method 2: The 8 Ô gets 16/(8 + 16) of total current. i8 = [(16)/(24)] * 6 i8 = 4.0 A

33. The current in the 15 ohm resistors is 0.37 A. What is the current in the 11 ohm?

The 11 Ô carries 7/(11 + 7) of total current i11 = [(7)/(18)] * .37 i11 = .144 A

34. What is the current through the 16 Ô?

iT = E/R iT = 12/[28 + 1/[(1/13) + (1/(16 + 20))]] iT = .320 A i16 = [13/(13 + 16 + 20)] * .320 i16 = .0849 A

If you find it natural and logical to think of the large current passing through the small resistor, you will find this shortcut helpful. If it just seems like one more equation, you'll do better to draw a potential diagram and use E = iR.

Another useful shortcut is the voltage divider. Here it is in all its glory.

35. What is the potential across the 24 Ô resistor?

i = E/R i = 9/(12 + 24) i = .250 A E24 = iR E24 = (.250)(24) E24 = 6.00 V note: The voltage divides in proportion to resistance: E24 = [24/(12 + 24)] * 9 E24 = 6.00 V

36. The current through the 50 Ô resistor is 0.83 A. What is the potential across the 22 Ô resistor?

E50 = iR E50 = (.83)(50) E50 = 41.5 V E22 = [22/(22 + 36)](41.5) E22 = 15.7 V

37. Find the current in the 7 Ô resistor by three different methods.

method 1: iT = E/R iT = 30/(8 + 1/[(1/12) + (1/7)]) iT = (30)/(12.42) iT = 2.42 A i7 = [12/(12 + 7)] * 2.42 i7 = 1.53 A method 2: From above, iT = 2.42 A E8 = iR E8 = (2.42)(8) E8 = 19.3 V ET = 30 - 19.3 ET = 10.7 V i7 = E/R i7 = (10.7)/(7) i7 = 1.53 A method 3: ET = E4.42 ET = [(4.42)/(8 + 4.42)](30) ET = 10.7 V i7 = E/R i7 = (10.7)/(7) i7 = 1.53 A

38. The potential across the 8 Ô resistor is 12 V. Find the following: i₈, i₁₁, V₂₄, i₂₄, i₄, V₄, V₇, i₃, V_T.

i8 = E/R i8 = (12)/(8) i8 = 1.5 A i11 = i8 i11 = 1.5 A V24 = V11, 8 V24 = iR V24 = (1.5)(19) V24 = 28.5 V i24 = E/R i24 = (28.5)/(24) i24 = 1.19 A i4 = i11 + i24 i4 = 1.50 + 1.19 i4 = 2.69 A V4 = iR V4 = (2.69)(4) V4 = 10.8 V V7 = iR V7 = iR V7 = (2.69)(7) V7 = 18.8 V i3 = [(9)/(9 + 3)] * 2.69 i3 = 2.02 A VT = V4 + V24 + V7 + V3 VT = 10.75 + 28.5 + 18.81 + (2.02 * 3) VT = 64.1 V

A battery is a source of energy, causing a current to flow. Where has this energy gone when the battery runs out? It can't simply disappear. Think about it it goes into heat. Wires get hot, resistors get very hot when current runs through them. The rate at which this energy is dissipated, the work per unit time, is called the power, just as it was in mechanics.

Recall:

power = (work)/(time)
power = charge * (potential)/(time)
power = current * potential

Thus we derive another equation for power.

P = iE (just remember "pie")

39. A 6 V battery delivers 2 amps. What is its power output?

P = iE P = (2 A)(6 V) P = 12 Watt note: amp * volt = (coul/sec)(J/coul) = J/sec = Watt

40. A 3 A current flows through a 6 Ô resistor. What power is dissipated?

E = iR E = (3)(6) E = 18 V P = iE P = (3)(18) P = 54 Watt

41. A 12 V battery is wired to a 24 Ô resistor. What power is dissipated by the resistor?

i = E/R i = (12)/(24) i = .50 A P = iE P = (.50)(12) P = 6.0 W

42. A current i flows through resistor R. What power is dissipated by it?

E = iR P = iE P = i(iR) P = i2R

43. A battery of potential E is connected to a resistor R. What power is dissipated by the resistor?

i = E/R P = iE P = (E/R)E P = E2/R

Note that we have two new expressions: P = i²R and P = E²/R as well as P = iE. Remember, these are not really new, they're just consequences of stuff we already know.

44*. The power output of the 3 Ô is 12 watts. What is the power output of the 5 Ô and the 2 Ô?

P = E2/R E2 = PR E2 = (12)(3) E = 6 V across the 3 Ô P5 = E2/R P5 = (62)/(5) P5 = 7.20 W i2 = iT i2 = E/R i2 = 6/(1/[(1/7) + (1/5)]) i2 = 3.2 A P2 = i2R P2 = (3.22)(2) P2 = 20.5 W

45*. A 30 V source is connected to a 20 Ô resistor submerged in 280 gm of water at 22°C. How long will it take to boil away all the water?

P = E2/R P = (302)/(20) P = 45 J/sec P * t = cmÇT + qfm 45T = (4200)(.28)(100 - 22) + (2.26 * 106)(.28) T = 1.61 * 104 note: All of the water will have boiled away after 4.5 hours

Electronic circuits often contain more than one circuit element, so we will finish the chapter by combining resistors with capacitors. Furthermore, circuitry is best mastered through repeated practice, as these final problems will indicate.

Capacitors normally appear in alternating current (AC) circuits, but some of their properties can be learned by thinking about them in direct current (DC) circuits. As AC circuits quickly get pretty difficult, we will confine our work to DC.

46. A 200 µF capacitor holds a charge of 6.0 * 10^-3C. It is discharged across the 4 Ô resistor by closing the switch. What is the current in the resistor immediately after the switch is closed?

i = E/R i = (q/c)/R i = [(6 * 10-3)/(200 * 10-6)]/(4) i = 7.50 A

47. Draw a graph showing how current in problem 46 changes as the capacitor discharges.

The initial potential on the capacitor is: V = q/c V = (6 * 10-3)/(200 * 10-6) V = 30 V Thus, the initial current is: i = E/R i = (30)/(4) i = 7.5 A As the capacitor discharges, the potential decreases so the current decreases, so the rate of drop of current lessens. Thus the graph will look like that on the right.

48. A 40 µF capacitor holding 350 µC of charge is discharged across a 150 Ô resistor. What is the initial current?

Initially the resistor receives the full potential of the charged capacitor. i = E/R i = (q/c)/R i = [(350 µC)/(40 µF)]/(150 Ô) i = 58.3 * 10-2 A

49. A 30 V potential is connected to the capacitor and resistor as shown. What is the initial current through the resistor? What is the current long after the circuit has been completed?

Initially the uncharged capacitor allows charge to flow onto it effortlessly: i = E/R i = (30)/(60) i = .50 A Within a few hundredths of a second the capacitor is essentially fully charged. Thus: i = 0

50. In the R-C circuit shown, what is the potential across the capacitor immediately as the switch is closed? What is it long after the switch was closed?

Initially there is no charge on the capacitor, so there can be no potential: V = 0 Eventually, when the capacitor is fully charged, the current flow stops, so there can be no potential across the resistor. Thus: V = 12 V

We see, then, that initially the uncharged capacitor in a circuit acts as a wire, transmitting current freely. When a little time has passed, the fully charged capacitor stops all current flow in its part of the circuit.

A study of the rate at which capacitors charge and discharge and their behavior in AC circuits requires calculus.

51. What is the initial resistance of this circuit? What is its resistance after equilibrium is reached?

Initially the capacitor allows current to completely bypass the 20 Ô, so: R = 50 Ô When fully charged, the capacitor ceases to conduct a current, so: R = 70 Ô

52. Find the current in the 12 Ô resistor (1) initially and (2) after a long time.

a) part 1: i = E/R i = (30)/(6 + 12) i = 1.67 A part 2: i = 0

b) part 1: i = E/R i = (30)/(12) i = 2.5 A part 2: i = 0

c) part 1: i = E/R i = (30)/(12) i = 2.5 A part 2: i = E/R i = (30)/(18) i = 1.67 A

53. Find the charge on each capacitor after equilibrium is reached.

a) V1 = (20/70)(12) V1 = 3.43 V q = CV q = (8 µ)(3.43) q = 27.4 µC

b) The 10 + 15 Ô resistors are irrelevant. .3 A flows through the 20 Ô and 40 Ô. E = iR E = (.3)(1/[(1/20) + (1/40)]) E = 4.0 V q = CV q = (14 µ)(4.0) q = 56.0 µC

c) V20 = (20/60)(50) V20 = 16.7 V V30 = (30/90)(50) V30 = 37.5 V So capacitors feel: V = 37.5 - 16.7 V = 20.8 V q = CV q = (11 µ)(20.8) q = 229 µC