My Problem below is, I have conflicting solutions from (i) the student solution

Question

My Problem below is, I have conflicting solutions from (i) the student solution manual and (ii) my professors solution given (from my assumption) the instructors solutions manual. Both approach (i) and (ii) ultimately achieve the same 'answer' but have two completely different approaches.

Both come from Physics, Cutnell, 8th edition (same ISBN).

Please Give me your thoughts on which solution is the "most correct".

Thanks!

Question:

Chapter 20 (Problem 20.99)

A 2.00-F and a 4.00-F capacitor are connected to a 60.0-V battery. What is the total charge supplied to the capacitors when they are wired

(a) in parallel and
(b) in series with each other?

Solution(s):

(i) From Student Solutions Manual: 

a. The Equivalent Capacitance In Parallel:

Ceq = C1 + C2

= 2.00 F + 4.00F

= 6.00 F  6.0 x 10-6 F

--> Total Charge Supplied To The Capacitor In Parallel:

QTotal = CeqV

= (6.00 F)(60.0-V)

= 36 x 10-5 C  3.6 x 10-4 C

 

b. The Equivalent Capacitance In Series:

--> Total Charge Supplied To The Capacitor In Parallel:

= 8.0 x 10-5 C

 

(ii) From Instructors Solutions Manual:

REASONING

       a.  When capacitors are wired in parallel, the total charge q supplied to them is the sum of the charges supplied to the individual capacitors, or q = q1 + q2. The individual charges can be obtained from q1 = C1V and q2 = C2V, since the capacitances, C1 and C2, and the voltage V are known.

 

       b.  When capacitors are wired in series, the voltage V across them is equal to the sum of the voltages across the individual capacitors, or V = V1 + V2. However, the charge q on each capacitor is the same. The individual voltages can be obtained from V1 = q/C1 and V2 = q/C2.

SOLUTION  

    a.  Substituting q1 = C1V and q2 = C2V  (Equation 19.8) into q = q1 + q2, we have:

 q = q1 + q2 = C1V1+C2V2 = (C1+C2)V = (2.00x10^-6F + 4.00x10^-6F)(60.0V) = 3.60x10^-4 C

   b.  Substituting V1 = q/C1 and V2 = q/C2  (Equation 19.8) into V = V1 + V2 gives

 V = V1 + V2 = q/C1 + q/C2

       Solving this relation for q, we have

 

______________________________________________

Explanation / Answer

Solution provided from student solution manual was the easiest one to understand clearly(which has circuit diagrams helps us to understand quickly). But you have to remember the following while answering the question. If capacitors are in parallel 1) The charge on each capacitor is same. 2) Equivalent capacitance C = C1 + C2 If capacitors are in series 1) Potential difference across the capacitors is same 2) Equivalent capacitance C = (C1C2)/(C1 + C2)               which comes from                                        1/C = 1/C1 + 1/C2

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