PHY-2049. Current & Circuits February ‘08. A closed circuit. Hot, Hot Hot. Power in DC Circuit. copper. 12 volts 0 volts.
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Current & Circuits
Hot, Hot Hot
12 volts 0 volts
The figure below gives the electrical potential V(x) along a copper wire carrying a uniform current, from a point at higher potential (x=0m) to a point at a lower potential (x=3m). The wire has a radius of 2.45 mm. What is the current in the wire?
What does the graph tell us??
*The length of the wire is 3 meters.
*The potential difference across the
wire is 12 m volts.
*The wire is uniform.
Let’s get rid of the mm radius and
convert it to area in square meters:
A=pr2 = 3.14159 x 2.452 x 10-6 m2
A=1.9 x 10-5 m 2
Material is Copper so resistivity is (from table) = 1.69 x 10-8 ohm meters
The rod in the figure is made of two materials. The figure is not drawn to scale. Each conductor has a square cross section 3.00 mm on a side. The first material has a resistivity of 4.00 × 10–3 Ω · m and is 25.0 cm long, while the second material has a resistivity of 6.00 × 10–3 Ω · m and is 40.0 cm long. What is the resistance between the ends of the rod?
In Fig. 28-39, find the equivalent resistance between points (a) F and H and [2.5](b) F and G. [3.13]
The ideal battery does work on charges moving them (inside) from a lower potential to one that is V higher.
This trip around the circuit is the same as a path through space.
THE CHANGE IN POTENTIAL FROM “a” AROUND THE CIRCUIT AND BACK TO “a” is ZERO!!
ADD ENOUGH RESISTORS, MAKING THEM SMALLER
AND YOU MODEL A CONTINUOUS VOLTAGE DROP.
Watch the DIRECTION !!
Let’s write the equations.
Resistors and Capacitors in the same circuit??
Is this cruel or what??
I need to use E for E
Note RC = (Volts/Amp)(Coul/Volt)
= Coul/(Coul/sec) = (1/sec)
qinitial=CE BIG SURPRISE! (Q=CV)
In Fig. (a), a R = 21, Ohma resistor is connected to a battery. Figure (b) shows the increase
Looking at the graph, we see that the
resistor dissipates 0.5 mJ in one second.
Therefore, the POWER =i2R=0.5 mW