Chapter 20 – Solutions to Problems 1, 5, 8, 11, 13, 17, 23, 25, 26, 30, 35, 37, 38, 39, 43, 47, 51

  1. A capacitor is a device. The capacitance of this device defines how much charge +Q and –Q, is stored in separated form for a given voltage difference applied to it.  The definition is C = Q/V.  A capacitor that stores 1 Coulomb of +charge on its positive “plate” and 1 Coulomb of –charge on its negative “plate” when a 1-volt difference is applied to its “terminals,” is said to have a capacitance of 1 Farad.  For the device described in this problem, Q = CV = (0.75 x 10-6 F)(25 V) = 19 microC.

  1. As described in problem 1), C = Q/V.  Thus C = 2 x 10-6 C)/(1000 V) = 2 x 10-9 F.

  1. If two concentric spheres have almost the same radii, then the gap between the two spheres is very small.  The two spherical surfaces can be regarded as closely equivalent to a parallel plate capacitor having the same area as that of a sphere having an intermediate radius.  This approximation becomes poor if the gap gets too large.  For this problem, the gap is only 0.50 mm, whereas the radii of the spheres are about 10 cm, so that the gap is only 5% of the average radius.  For a large gap you would need calculus, which is beyond expectation for this course.  Recall that the surface of a sphere of radius R is 4(pi)R2.  Your book tells you that the capacitance of a parallel-plate capacitor is C = (epsilon0)(A)/d.  Thus C is approximately equal to (8.85 x 10-12 F/m)(4)(3.14)(0.105 m)2/(0.0005 m) = 2.5 x 10-9 F.

  1. In class, we discussed that the effective capacitance of three capacitors in parallel is found from the expression Ceff = C 1 +C2 + C3, and of three capacitors is series from the expression

    .

    1. Thus for the three capacitors in series 1/Ceff = (1/2 + 1/7 + 1/14) microF-1 = 10/14 microF-1.

                Ceff = 1.4 microF.

    2. In parallel, the three capacitors are simply added to get the effective capacitance.

                Ceff = (2 + 7 + 14) microF = 23 microF.

  1.  
    1. Remember the important comments in class about various voltages across circuit elements in any circuit.  When three capacitors are connected in parallel across a 12-volt battery, each capacitor has 12 volts across it.  For each capacitor, Q = CV, so that

                      Q2 = (2 microF)(12 V) = 24 microC

                      Q6 = (6 microF)(12 V) = 72 microC

                      Q12 = (12 microF)(12 V) = 144 microC

    2. The series configuration is not that simple, however.  Refer to the diagram below.  The important starting point is to know that each capacitor in the series has the same stored charge on their individual positive plates.  Thus replace the series by its effective value using the series equation of problem 11.  Since 1/2 + 1/6 + 1/12 = 9/12, we see that the effective capacitance if the series is Ceff = 4/3 microF.  This effective capacitor, when charged by 12 volts, stores a charge of

                Q = (4/3 microF)(12 V) = 16 microC.

      A charge +Q resides on the positive plate, and a like amount –Q resides on the negative plate of each capacitor.  This fact is all we need to determine the voltage across each capacitor.

                VAB = VB – VA = – (16 microC)/(2 microF) = –8 V.

                VBC = VC – VB = – (16 microF)/(6 microF) = –2.67 V

                VCD = VD – VC = – (16 microF)/(12 microF) = –1.33 V

  1. The solution of this question is based on the concepts of Chapter 19.  To begin with, the electric field between the plates of a parallel plate capacitor is uniform, except at the edges.  In a uniform electric field, the voltage difference between any two points along the field is Ed.  The imaginary plane perpendicular to the electric field is called an equipotential surface, because everywhere on this plane, the electrical potential is constant.  Two such perpendicular planes 10 cm apart in a field of strength 30,000 V/m (remember it’s also 30,000N/C) is assumed to have a potential difference of 10 V.  Thus, the separation distance must be

                              d = (10 V)/(30,000 V/m) = 3.33 x 10-4 m.

  1. Current is defined as I = the amount of charge passing a point each second.  If the charge passing a point is 1 Coulomb each second, the current is defined as 1 Ampere (A).  Therefore a current of 0.5 A must represent a charge flow of 0.5 C per second.  Also, since the charge of a single electron has a magnitude of 1.6 x 10-19 C, 0.50 C must be

                                    (0.5 C)/(1.6 x 10-19 C per electron) = 3.1 x 1018 electrons.

  1. As stated in the last problem, the current = (the charge passed)/(time).  Thus,

                                    I = (1 x 10-9 C)/(1 x 10-7 s) = 0.01 A.

  2. A resistor is a circuit element that limits how much current passes through when a voltage difference is applied across it.  Ohm’s law defines resistance through the relation V = IR, where the proportionality constant R is called the resistance.  If 1 A flows when a voltage drop of 1 V is applied to a resistor the element is said to have a resistance of 1 ohm.   “Ohms” is expressed by the Greek Letter “capital omega,” but since some browsers do not support Greek letters I will always write out the word ohms.  In this problem, a current of 3 A flows through a 12-ohm resistor.  Thus,

                                    V = IR = (3 A)(12 ohms) = 36 V.

  1. Notice that the rules by which resistors in combination are evaluated to obtain equivalent resistance is opposite in form when compared with capacitors.  Assume the three resistors 5, 7 and 9 ohms. 

    1. In series, total resistance is the sum of the three individual resistances.

      Reff = 5 ohms + 7 ohms + 9 ohms = 21 ohms.

    2. In parallel the effective resistance is found from 1/Reff = (1/5 + 1/7 + 1/9) ohms-1.  Evaluating this sum by using the common denominator method gives

      1/Reff = 143/315 ohms-1 or Reff = 2.2 ohms.

  1. The 5-ohm and 4-ohm resistors are in parallel.  These two should be combined first to get an “first step” value of resistance.  I will call it R’

    1/R’ = (1/5 + 1/4) ohms-1 or R’ = (20/9) ohms = 2.2 ohms.

    Then this equivalent resistance of the parallel pair is in series with the 3-ohm resistor.  Therefore,

    Reff = R’ + 3 ohms = (2.2 + 3.0) ohms = 5.2 ohms.

  1. The first step is to observe that there are no resistors in parallel but that the 2- and 5-ohm resistors are in series.  This effective 7-ohm resistor is in parallel with the 6-ohm resistor.  Thus, the circuit has been reduced partially to the circuit shown, where R1 is found from 1/R1 = 1/7 + 1/6 to give R1 = 3.23 ohms.

         

    The next step is to join R1 with its series partner of 3 ohms, this resulting effective 6.23-ohm resistor is in parallel with the 10-ohm resistor.  The final result is Reff = 3.84 ohms.

  2. Recall from class how this was redrawn to make it look more familiar as a combination of

    1. a 3- a 6- and a 5-ohm resistor in parallel joined to

    2. a 4- and a 10-ohm pair of resistors in parallel joined to

    3. a 2-ohm resistor.

    Using the formulas for parallel resistors, we find the effective values 1.43 and 2.86 ohms.  Adding up the three series resistors gives 1.43 + 2.86 + 2.0 = 6.29 ohms.

  3. This is an important problem to understand very clearly. 

    1. The answer to problem 20-35 was Reff = 5.22 ohms.  This effective resistance is what the 12-volt battery sees.  As a result, the current through the battery is I = (12 V)/(5.22 ohms) = 2.30 A. 

    2. This current of 2.30 A goes through the 3-ohm resistor, then splits up with part of it going through the 4-ohm resistor and the rest going through the 5-ohm resistor.  Here is where the misunderstandings begin.  The battery provides a voltage rise of 12 V.  But the voltage drop across the 3-ohm resistor must be V = IR = (2.30 A)(3 ohms) = 6.9 V.  Now remember that the sum of voltages around any circuit loop must be zero.   Therefore, the voltage drop across the 4-ohm resistor must be 12 – 6.9 = 5.1 volts.  This information then tells us what we nee to find the current through the 4-ohm resistor.  The answer is I = V/R = (5.1 V)/(4 ohms) = 1.28 A.

  1. Unfortunately, this problem is worded differently in different printings of the book.  My book says “How much power is lost in a resistor when a 3.0-V battery sends a current of 0.30 A through it?”  The problem is a very simple one.  The power is given by P = IV. 

    P =  (0.30 A)(3.0 V) = 0.90 watts.

  1.  

    1. At 220 volts, the power expended is (220 V)(7.5 A) = 1650 watts

    2. t 120 volts the same power as before requires a current of

      I = P/V = (1650 W)/(120 V) = 13.8 A.

    Because high current causes excessive heating in house wires, appliances that need large power are usually operated at 220 volts so that current does not need to be as high.

  1. The equation relating resistance to resistivity is R = (rho)(L)/A.  For #10 Nichrome wire having a diameter d = 2.59 x 10-3 m (radius is 1.295 x 10-3 m) and a resistance of R/L = 0.213 ohms/m, The resistivity is

    rho = (R/L)(Area) = (0.213 ohms/m)(3.14)(1.295 x 10-3 m)2 = 1.12 x 10-6 ohm.m.