Jaquin Spring 2011
PHY 104 Exam #2
Answer all the questions that appear below.
1)A 5.00 pF parallel plate capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials up to 200 V. The electric field between the plates is no to exceed 50,000 N/C.
A)What should the separation of the plates be?
Use the relationship between Voltage and Electric Fields for parallel plate capacitors. The work done by the field to move a test charge q across the plates is qV. Therefore, and . The plates should be separated by just 4 mm.
B)What charge will the plates hold at maximum potential?
Use the definition of capacitance and solve for QMax. The charge on the capacitor at maximum potential is .
2)What is the net capacitance of the following capacitor network?
First collapse the two capacitors in parallel using
Next collapse the series of three capacitors using. The value of the CEQ works out to be 1.39 F.
3)A copper wire at 20C has a diameter of 2.54 mm , a length of 25 cm and carries 5.00 A of current. How many electrons pass any given point in the wire over a 4 second period?
There is much more information given in this problem that what is needed. WWe only need to know the current.
By definition . Therefore the charge that passes past some point in the wire after 4 seconds is given by.
Thus electrons pass any given point in the wire over a 4 second period.
4)A lead cylinder with length L and radius r has a resistance of R. The cylinder is dropped from a great height onto a solid surface and compresses its length by 15%. Assuming that the deformed lead is still has a cylindrical shape; by what percentage did the resistance change?
The picture to the right illustrates the shape of the cylinder before and after dropping. The volume of the cylinder, V=L∙A, remains unchanged. Therefore, L∙A= L'∙A'. Thus.
The resistance of the cylinder across its ends is given by before the fall and after the fall. Examining the resistance after the fall R', we see that
Thus the resistance of the cylinder decreases by 28% or is 72% of its original resistance.
5)26 gauge copper magnet wire has a diameter of 0.4 mm. What length of this wire is needed to have a resistance of 25 Ohms at room temperature (20C)?
Use the relation for resistance and solve for L.
The copper wire would need to be 187 m long to have a resistance of 25 Ohms.
6)What would be the resistance of this copper wire if it were cooled to liquid nitrogen temperatures of -200C?
We can use the temperature dependent relation for resistance, , where is the temperature coefficient of resistivity in units of C-1. For copper the temperature coefficient of resistivity is 0.0039 C-1. So the new resistance will be
The resistance of the copper wire drops to 3.55 Ohms at -200 Celsius.
7)With a 1,500 M resistor across its terminals, the terminal voltage of a certain battery is 2.50 V. With only a 5.00 resistor across its terminals, the terminal voltage is 1.75 V. What is the internal emf and the internal resistance of this battery?
a)This is a bit of a trick question in that the 1,500 M (1x109) resistor acts essentially like an open circuit (i.e. infinite resistance). Thus the internal emf of the battery is just 2.50 V since there is no appreciable internal resistance loss due to the astronomically high 1,500 M resistor.
b)The internal resistance can be solved if you know the terminal emf (1.75 v) and the current through the battery using E- Ir = Terminal Voltage, where E is the internal emf of the battery and r is the internal resistance of the battery. The potential between point s 1 & 2 (a.k.a. the terminal voltage) in the circuit as shown in the diagram is 1.75 V. Therefore, from Ohm’s Law, there is I = V/R = (1.75 v)/ (5 ) = 0.35 A flowing through the resistor and also flowing through the battery. Thus we can solve for the internal resistance
Thus the internal resistance of the battery is 2.14 Ohms.
8)A toaster using a NichromeTM heating element operates on 120 V. When it is switched on it will draw a steady current of 1.4 A. How much energy is consumed by the Toaster over a toasting cycle of 1.5 minutes?
The energy is consumed by the Toaster over a toasting cycle of 1.5 minutes is 15,120 Joules.
9)Calculate the equivalent resistance of the following resistance network.
First, re-draw the circuit so as to resemble the circuit at the left. Then combine the two resistors in parallel. Their equivalent resistance is 10 .
Next, combine the two pairs of series resistors. The two on top will have an equivalent resistance of 30 and the two on the bottom will have an equivalent resistance of 40 .
Next, combine the two new parallel resistors. Their equivalent resistance will be 17.14
Finally, combine the three series resistors into a single equivalent resistance of 57.14
10)Calculate the current flowing through the 4 Ω resistor given that the current through the 1 Ω resistor is 3 A as shown, the current through the 2 Ω resistor is -3 A as shown, and the battery supplies 90 C of charge every 20 seconds as indicated.
This problem is much simpler than it first appears. The key is to simply solving this is to interpret the statement “the battery supplies 90 C of charge every 20 seconds as indicated”. This means that the battery has a current of flowing through it in the upward direction indicated in the figure. Thus the currents at the junction just below the 2 resistor are as shown below in red. The unknown current must be 7.5 A according to Kirchhoff’s Junction Rule. This is also the current through the 4 resistor.
11)Determine the current through each of the resistors in the circuit shown below and identify the resistor that dissipates the most power.
V1 / 10 VV2 / 6 V
R1 / 10 Ω
R2 / 20 Ω
R3 / 2 Ω
R4 / 8 Ω
This is a full blown Kirchhoff’s Loop Law problem. The loop equations for the current directions I have drawn above appear below.
Redistributing terms…
Substituting in values and dropping units …
I obtained solutions of the two equations for I1 and I2 of …
The current through each resistor is then…
Resistor / Current / Power (I2R)R1=10 / 0.04 A to Right / 0.016 W
R2=20 / 0.40 A to Right / 3.20 W
R3=2 / 0.36 A to Left / 0.26 W
R4=8 / 0.36 A Down / 1.04 W
The R2 resistor draws the most power.
12)An uncharged 5 µF capacitor is connected in series with a 50 Ω resistor and a 12.0 V battery.
A)What is the time constant of the RC circuit?
The time constant is given by .
B)Sketch the graphs of current in the circuit and the charge on the capacitor as functions of time.
C)What will be the charge on the capacitor 0.5 ms after the charging begins?
The governing equation is . The maximum charge, which would be obtained only after a long time (greater than 10 time constants), is calculated from
.
Thus at t = 0.5 ms, the charge on the capacitor is
Thus the capacitor has 51.6 C of charge or 86% of its final charge after 0.5 ms.
D)What will be the potential across the resistor at 0.5 ms after charging begins?
Use Ohm’s Law, V=IR, to calculate the potential across the resistor. However, first the current must be found using the governing equation for RC circuits. The constant I0is the maximum current that is experienced immediately before the capacitor has time to charge.
Thus at t = 0.5 ms, the currentthrough the resistor is
Thus the potential across the resistor is32.5 mA or 13.5% of its initialmaximum value after 0.5 ms.