Tuesday Lecture, Aut. 2001; the physical  is real !


1. The capacitors have values C1 = 10.0 m F , C2 = 20.0 m F , C3 = 20.0 m F , C4 = 40.0 m F, and C5 = 10.0 m F . The battery has value 20.0 Volts. Note : 1 m F = 10-6 F.

(a) ( 30 points) What is the total equivalent capacitance of the circuit ?
(b) (6 points) If the distance d between the parallel plates for C5 is d = 0.002 m, then what is the magnitude E of the electric field between those plates ?
(c) (2 points Extra Credit) What is the magnitude of the charge on the positive plate for C4 ?
(d) (1 points Extra Credit) If the distance d' between the parallel plates for C4 is d' = 0.002 m, then what is the magnitude E of the electric field between those plates ?
 

Solutions:
(a) 1/10  +  1/20  = 1/Cs1  so Cs1 = 20/3  and 1/ 20 + 1/40 = 1/Cs2  so that Cs2 = 40/3  so that Cs1 + Cs2 = 60/3 = 20 uF. Thus, 
1/Ctot = 1/20 + 1/10   so that Ctot = 20/3 uF = 6.67 uF.
(b) Ed = V5   where V5 = Q5/C5   but Q5 = Ctot*20  =133.4 uF  
E = 133 uF/(C5*d) = 133.4 x10-6/(1x10-5*2x10-3) =6.67 x10 3 V/C
(c) Q4 = Cs2*(voltage across Cs2)= Cs2*(20 - Q5/C5)=
Cs2*(20 - 40/3)= (40/3)(20/3) uC = 800/9 uC =88.8 uC
(d)  Ed = Q4/C4 = 88.8 uC/ 40 uF = 2.22 Volts so that E = 2.22/2x10-3  = 1.11x103 V/C

2. (30 points) The heating element of a coffee maker operates at V = 120.0 Volts and carries a current of i = 2.00 A. Assume all the heat generated by the resistor is absorbed by water of mass 0.500 kg. How long (in seconds) does it take to heat the water from exactly 23 0C to the boiling point of exactly 100 0C and then vaporize 0.0500 kg of the water into steam? Cw = 4186 J/oC kg and Lv = 2.26x106 J/kg

Solutions:
iVt  =  mLv  +  mC(100- 23)
2*120*t = 0.5*2260000  + 0.5*4186*77 so that t = 1142.3 seconds

 3. Extra Credit. (5 points) The battery voltage = 10 Volts. Assume:

(a) ( 3 points) Find the current in resistor R3 .
(b)  (2 points) Find the current in resistor R1.

Solutions:
Do (b) first:
(b) 1/ Rp  = 1/2  + 1/1   + 1/4 so Rp = 4/7  so
Rtot = 1 + 1 + 4/7 = 2.57 ohms  so that itot = 10/2.57 = 3.89 A = current in R1
(a)  i3 = itot*Rp/R3   = 3.89*(4/7)/1 = 2.22 A