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BW 8 Ch. 23: CQ 17 – 22 P. 32 – 36
Q23.17. Reason: Applying Kirchhoff’s loop law around the outside edge of the circuit,
That is, the potential difference DV12 between points 1 and 2 is potential supplied by the battery minus the potential lost
in resistor R. When bulb B is in place, a current I exists through the resistor and the bulb. In that case, DV12 is less
than
But if bulb B is removed the current no longer exists in that branch (I = 0 A).Thus
The
answer is that the potential difference DV12 increases.
Assess: Removing bulb B means that there is no resistor between points 1 and 2. “No resistor” is not the same
as “no resistance.” “No resistor” means an insulator with R = ¥ W, not R = 0 W.
Ohm’s laws applies to resistors, not to empty spaces. Although I = 0A when bulb B is removed, you cannot use
IR to conclude that DV12 = 0 V because there is no longer a resistor between points 1 and 2 to which Ohm’s law
could be applied.
Q23.18. Reason: (a) The total current splits and some goes through the parallel branch containing the point a
(the left branch) and some goes through the parallel branch containing the point b (the right branch). Notice that
the resistance of each branch is the same (3R), as a result the current splits equally between the two branches.
Let’s call the current in the two branches I. This current in the two branches causes a potential drop of 2IR in the
resistor 2R in the left branch and a potential drop of IR in the resistor R in the right branch. As a result, the
electric potential at point b is greater (it is more positive) than the electric potential at point a and DVab ¹ 0.
(b) Since point b is at a higher potential than point a, if they are connected there will be an electric current from
point b to point a.
Assess: The solution of the question requires a good understanding of electric potential difference.
Q23.19. Reason: (a) When the bulb is removed the resistance of that branch becomes ¥.The equivalent
resistance of the parallel combination increases and is the resistance of the single -R branch. Since the resistance
of the parallel combination has increased, the current decreases.
(b) When the bulb is removed there is no potential difference across the resistor that is in series with the bulb
because there is no current through the resistor (DV = IR). So the potential difference between 1 and 2 is now the
potential difference of the parallel group. That is, removing the bulb increases the potential difference between 1
and 2.
Assess: The second part is like Question 22.17, and we arrived at the same answer here.
Q23.20. Reason:
(a) A good voltmeter has very high resistance which causes the current in the incorrect circuit to be nearly zero.
(b) Because there is no current there is no potential difference (voltage drop) across the resistor, so the voltmeter
measures the emf of the battery, or 9 V.
(c) Put the voltmeter in parallel with the resistor rather than in series with it.
Assess: Voltmeters should be in parallel with the circuit element we want to know the potential difference across.
Q23.21. Reason:
(a) Because a good ammeter has very low resistance, the current would nearly all go through the ammeter, so the
current in the 5.0 W resistor is nearly zero.
(b) Put the ammeter in series with the resistor rather than in parallel with it. The current will be the same
anywhere in the one-loop circuit, so it doesn’t matter exactly where the ammeter is placed in series.
Assess: Ammeters should be placed in series in the circuit where we want to know the current.
Q23.22. Reason: The equivalent capacitance of each grouping is determined as follows:
Group 1: In this group we have three capacitors in series and the equivalent capacitance is 1/Ceq = 1/C1 + 1/C2 +
1/C3 or Ceq = C/3.
Group 2: In this group we have three capacitors in parallel and the equivalent capacitance is
Ceq = C1 + C2 + C3 = 3C.
Group 3: In this group we have two capacitors in parallel and this combination is in series with the third capacitor. First
determine the equivalent capacitance of the parallel combination and then consider this equivalent capacitance to be in
series with the third capacitor. The equivalent parallel capacitance is CParallel = C + C = 2C. The equivalent capacitance
for the group is determined by 1/Ceq = 1/CParallel +1/C = 1/(2C) +1/C = 3/(2C) or Ceq = 2C/3.
Group 4: In this group we have two capacitors in series and this combination is in parallel with the third capacitor. First
determine the equivalent capacitance of the series combination and then consider this equivalent capacitance to be in
parallel with the third capacitor. The equivalent series capacitance is 1/Cseries = 1/C +1/C = 2/C or CSeries = C/2. The
equivalent capacitance for the group is determined by Ceq = Cseries + C = C/2 + C = 3C/2.
This allows us to rank the capacitance of the groups as follows: C2 = 3C > C4 = 3C/2 > C3 = 2C/3 > C1 = C/3.
Assess: The solution of this question requires a good knowledge of how capacitors add in parallel and in series.
P23.32. Prepare: Capacitors in parallel follow Equation 23.17.
Solve:
The equivalent capacitance is
Ceq  C1  C2  C3  6.0 F  10 F  16 F  32 F
Assess: This value is the sum of all capacitances, so the result is reasonable.
P23.33. Prepare: Capacitors in series follow Equation 23.19.
Solve:
The equivalent capacitance is
-1
-1
æ 1
æ 1
1
1ö
1
1 ö
Ceq = ç +
+ ÷ =ç
+
+
= 3.04 ´ 10-6 F = 3.0 m F
è 6.0 m F 10 m F 16 m F ÷ø
è C1 C2 C3 ø
Assess: This value is the smallest of all capacitances, so the result is reasonable.
P23.34. Prepare: Two capacitors in parallel combine to give greater capacitance according to Equation
23.17.
Solve: Since we want a capacitance of 50 F and we have a 30 F capacitor, we must connect the second
capacitor in parallel with the 30 F capacitor. That is,
C  30 F  50 F Þ C  50 F  30 F  20 F
Assess: We must learn how the series and parallel circumstances combine.
P23.35. Prepare: Two capacitors in series combine to give less capacitance according to Equation 23.19.
Solve: Since we have a 75 F capacitor and we want a 50 F capacitance, we must connect the second
capacitor in series with the 75 F capacitor. The capacitance of the second capacitor is calculated as follows:
1
1
1
Þ C  150 F
+ =
75 m F C 50 m F
Assess: We must learn how to combine series and parallel capacitances.
P23.36. Prepare: Please refer to figure P23.36. The pictorial representation shows how to find the
equivalent capacitance of the three capacitors shown in the figure.
Solve: Because C1 and C2 are in series, their equivalent capacitance Ceq 12 is
1
Ceq 12
=
1
1
1
1
1
Þ Ceq 12  12 F
+
=
+
=
C1 C2 20 m F 30 m F 12 m F
Then, Ceq 12 and C3 are in parallel. So,
Ceq  Ceq 12  C3  12 F  25 F  37 F
Assess: We must understand well how to combine series and parallel capacitance.