Quiz  19 (ANSWERS)  1, 2, 3, 4, 5, 6, 14, 17, 18, 19, 20, 27, 28, 37, 38, 41, 44,  50, 51
TURN IN: 2, 4, 6, 14, 17 (Find the current in each resistor ), 18, 28, 38, 44, 50

2,4,6 were reviewed in class. On 6 (b),  just remember that if resistors are in  parallel, you add the reciprocals of them to get the net parallel resistance. Thus
 1/RP = 1/R1  + 1/R2  + 1/R3  + 1/R4  + 1/R5 + 1/R6.  Evaluate the inverse of the right hand side to find RP. 

14. Assume the light  bulbs are identical. Find the equivalent parallel resistance of the light bulbs from the formula:
1/Rp = 1/R  + 1/R  + 1/R  + 1/R  + 1/R + 1/R  + 1/R + 1/R
Then find the total equivalent resistance by by evaluating the sum:
Rp + 1.6 ohm 
17. This was reviewed  during lecture; I will provide the discussion to the entire class now.
Find the equivalent parallel resistance of the light bulbs from the formula:
1/Rp = 1/(820 ohm)  + 1/(680 ohm) .

Then find the total equivalent resistance by evaluating the sum:
RTOT = Rp + 470 ohm.

Once you find the total resistance, find the total current flowing through the battery and equivalent resistance using I = 12 (V)/RTOT   .

Now you are in a position to answer questions ((b) and (c).

(b) The voltage across each  the 820-ohm and 680-ohm resistors are the same; see page 523. That common voltage drop has magnitude I*RP.  The voltage drop across the 470-ohm resistor has magnitude
I*(470 ohm).
(c) The current though the 470- ohm resistor is the I computed  above. The current through the parallel
820-ohm and 680-ohm resistors are easy to compute. Just divide their common voltage drop magnitude (see  part (b) ) by  820 ohm and 680 ohm, respectively.

18. Power (in Watts) = V2/R. The power and the voltage drop magnitude V are given for each light bulb.  Find R for each bulb, then add them to find the total equivalent resistance of the in-series resistors .
28. See example 19.8 and my class notes.  In series with each battery, insert in the diagram resistors of magnitude r = 1.2 ohm. The diagram should look something like Figure 19-13 where we see 1 ohm resistors in series with each of the  batteries. You will have three equations and three unknowns as in example 19.8. HOWEVER WE DISCOVERED THE  PROBLEM IS EASY; YOU CAN IMMEDIATELY GET THE VOLTAGE DROP  ACROSS R2,  WHICH IS 16.2 OHMS  WHEN YOU INCLUDE THE INTERNAL RESISTANCE  OF 1.2 ohms.  YOU CAN IMMEDIATELY GET I2 THROUGH R2. Once you get I2, get I1 using Kirchkoff's rule for the sum of the voltages around a loop----in this case, go clockwise around the  top loop:
- 9.0 (V) + I2*R2 - I1*R1 = 0.
38. See section 19.5.  See examples 19.10 and 19.11.  See class notes. 

The voltage across the 2.00 uF capacitor is 26.0 (V) because  it is in parallel with the  battery. 
Note: 1uF = 10-6   F.  

You should now find the equivalent capacitance  of the two series connected capacitors  from the formula:
1/Cs = 1/C1  + 1/C2   , where C1 = 3.00 uF and C2 = 4.00 uF. Note: 1uF = 10-6   F. 
Compute Cs then note that Cs = Q/V, where V = 26.0 (V)  since Cs is in parallel with the battery. Compute  Q.  Note that Q is the common charge shared by the positive plates of C1 and C2.  Evaluate V1 = Q/C1 and
V2 = Q/C2.
44.  See #38 hint  above.  Find the charge on each capacitor. The method for finding  the common Q for C2 and  C3  is shown there.  To find the charge Q1 on C1 just use  C1 = Q1/V.  Once you  find all the charges, find the energy for each capacitor using equations 17-10. Then add them up to get the total network energy.
50. Read section 19-6.
(a) See equation 19.7.
(b) This will be discussed in class.