Physics 202 Homework, Day 13: Chapter 18: #2,4,6a,7,19
ü 2. Three 9.0 W resistors are connected in series with a 12-V battery. (a) Find the equivalent resistance of the
circuit, and (b) the current in each resistor. (c) Repeat for the case in which all three resistors are
connected in parallel across the battery.
„ Resistors in series
In[27]:=
rSeries = 3. + 3. + 3.; Print@"Series Resistance = ", rSeries, " Ohms"D
12
seriesCurrent =
; Print@"Current = ", seriesCurrent, " Amperes"D
rSeries
Series Resistance = 9. Ohms
Current = 1.33333 Amperes
„ Resistors in parallel
In[31]:=
rParallel =
1
1
3.
+
1
3.
parallelCurrent =
+
1
; Print@"Parallel Resistance = ", rParallel, " Ohms"D
3.
12
rParallel
; Print@"Current = ", parallelCurrent, " Amperes"D
Parallel Resistance = 1. Ohms
Current = 12. Amperes
ü 4. (a) Find the current in a 8.00 W resistor conected to a battery that has an internal resistance of 0.25 W if
the voltage across the battery terminals is 9.00 V.
(b) What is the emf of the battery?
In[44]:=
internalResistance = 0.25; totalResistance = 8. + internalResistance;
9.00
current =
;
totalResistance
Print@"Current = ", current, " Amperes"D
Current = 1.09091 Amperes
9.00 V = (emf) - (current)*(internal resistance) => emf = 9.00 V + (current)*(internal resistance)
In[48]:=
emf = 9.00 + current * internalResistance;
Print@"emf = ", emf, " Volts"D
emf = 9.27273 Volts
ü 6a. Consider the combination of resistors shown in the figure below. (a) Find the equivalent resistance
between point a and b.
2
Day_13_sol.nb
6a. Consider the combination of resistors shown in the figure below. (a) Find the equivalent resistance
between point a and b.
In[10]:=
rParallel1 =
1
1
12.
rParallel2 =
+
6.
1
1
4.
+
;
1
1
;
8.
rTotal = rParallel1 + 5. + rParallel2; Print@"Total resistance = ", rTotal, " Ohm"D
Total resistance = 11.6667 Ohm
ü 7. What is the equivalent resistance of the combination of identical resistors between points a and b?
Express your answer interms of R.
The three resistors in parallel have an equivalent resistance of
The total resistance is R + R/3 + R = (7/3) R.
ü
@1 ê R + 1 ê R + 1 ê RD-1 = R / 3.
Day_13_sol.nb
ü 19. The figure below shows a circuit diagram. Determine (a) the current, (b) the potential of the wire A
relative to ground, and (c) the voltage drop across the 1500 W resistor.
(a) Use the Loop Rule (Kirchhoff's Voltage Law), choosing the clockwise direction and starting at the lower left.
20 - I(2000W) - 30 - i(1000W) => 15 = I (2000W + 1000W +
=> I = 15 / (5000W) = 3.0 mA
In[56]:=
current =
15.
5000.
I(1500W) + 25 - I(500W) = 0
1500W + 500W) = i (5000W)
; Print@"Current = ", current * 1000, " mA"D
Current = 3. mA
(b) Use the Loop Rule (Kirchhoff's Voltage Law), choosing the clockwise direction and starting at the lower left, this time
stopping
at
the
top
of
the
loop
(point
A):
20 - I(2000W) - 30 - i(1000W) = Voltage at A
In[58]:=
Print@"Voltage at A = ", 20 - current * 2000 - 30 - current * 1000, " Volts"D
Voltage at A = - 19. Volts
(c) Voltage drop across the 1500W resistor = (3 x 10-3 A)(1500W) = 4.5 Volt
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