# Electricity Problems: Understanding Temperature Coefficient and Resistivity

• teb9186
In summary, for a long underground cable with two parallel wires and a short circuit at distance x from the west end, the resistance of the wires and the short is given by RE = rL(L-x) + R = 120 W and RW = rLx + R = 200 W. Solving algebraically, we can find that x = (RE - RW)/2rL and R = (RE + RW)/2. Plugging in the given values, we get x = -4 km and R = 160 W.
teb9186
Here's the problem:
The temperature coefficient of resistivity alpha is given by alpha = dr/rdT where r is the resistivity at temperature T. From this expression follows r(T) = ro[1 + a(T - To)], if a is presumed to be constant and much smaller than (T - To)-1.

(a) (3 points) However, if a is not constant, but is given by alpha = -n/T, show that r = a/T^n^, where n is a dimensionless constant, r is given in W·m and T is given in kelvin (K). Such a relation might be used as a rough approximation for the current temperture dependence of the resistivity for a semiconductor.

(b) (1 point) Using the values r = 3.50 × 10-5 W·m and alpha = -5.00 × 10-4 K-1 for
graphite at 293 K, (note: the w's are ohm symbols)
determine a

The first thing I did was differentiated r(t) and divided that by r and set that equal to alpha. I got alpha=ro(T-To)/r since this equals -n/T i set the two equal and found that r=(-ro9T-To)T/n
I don't know how to get what I got for r into the form given. Can somebody help?

I also have this short circuit problem and I have no idea where to started:
A long underground cable with length L = 12.0 km extends east to west and it consists of two parallel wires, each of which has a linear resistivity rL = 10.0 W/km. A short develops at distance x from the west end when a conducting path of unknown resistance R connects the wires. (See the figure above.) The resistance of the wires and the short is then RE = 120 W when the measurement is made from the east end, and RW = 200 W when it is made from the west end.
(a) (4 points) Find x and R algebraically in terms of L and rL, RE and RW.
(b) (1 point) Evaluate your results numerically.
Again the w's are ohms

I have been trying different things but I am very confused. I don't understand how a short circuit would happen in this situation. Arent the two wires that are connected at the same potential, meaning know charge would flow between them?

(a) To find x and R, we can use the fact that the total resistance of the circuit is equal to the sum of the individual resistances. So, we have:

RE = rL(L-x) + R
RW = rLx + R

Solving for R in both equations, we get:

R = RE - rL(L-x)
R = RW - rLx

Setting these two equations equal to each other, we get:

RE - rL(L-x) = RW - rLx

Solving for x, we get:

x = (RE - RW)/(rL + rL) = (RE - RW)/2rL

Substituting this value of x into either of the original equations for R, we get:

R = RE - rL(L - (RE - RW)/2rL) = (RE + RW)/2

(b) Plugging in the given values, we get:

x = (120 - 200)/(2*10) = -4 km (since the value is negative, this means the short circuit is located 4 km from the west end)
R = (120 + 200)/2 = 160 W

I can provide a response to the content presented.

First, for problem (a), we can start by rearranging the given expression for alpha to solve for r. This gives us r = ro[1 + a(T-To)]. Then, we can substitute the given expression for a, which is alpha = -n/T. This gives us r = ro[1 - n(T-To)/T]. Finally, we can simplify this expression to get r = ro(T^n^/T^n^-n(T-To)). This can also be written as r = a/T^n^, where a = ro(T^n^-n(T-To)). Therefore, we have shown that r = a/T^n^, where n is a dimensionless constant.

For problem (b), we can plug in the given values for r and alpha into the expression we found in part (a). This gives us 3.50 × 10-5 W·m = a/(293 K)^n^. We can then solve for a by multiplying both sides by (293 K)^n^ and we get a = 3.50 × 10-5 W·m × (293 K)^n^.

Moving on to the short circuit problem, we can approach it by setting up a system of equations. We know that the total resistance of the circuit is equal to the sum of the resistances of the two wires (RW + RE) and the resistance of the short (R). This can be written as RW + RE + R = Total resistance. We also know that the resistance of each wire is given by rL * L, where L is the length of the wire. Therefore, the total resistance can also be written as 2rL * L + R = Total resistance.

For part (a), we can solve for R by subtracting 2rL * L from both sides and we get R = Total resistance - 2rL * L. We can then substitute in the given values for RW and RE to get R = 120 W - 2(10.0 W/km)(12.0 km) = 0 W. This means that the resistance of the short circuit is 0 W.

To find x, we can use the fact that the total resistance is equal to the sum of the resistances of the two wires and the short circuit. This can be written as 2rL * x + R = Total resistance

## What is electricity?

Electricity is the flow of electric charge through a conductor, such as a wire. It is a form of energy that powers many devices and is essential for our daily lives.

## What are the common causes of electricity problems?

Some common causes of electricity problems include faulty wiring, power surges, overloading circuits, and outdated electrical systems. Extreme weather conditions and natural disasters can also cause electricity problems.

## How can I troubleshoot electricity problems in my home?

If you are experiencing electricity problems in your home, first check to see if any circuit breakers have tripped or fuses have blown. If not, try unplugging any devices that may be causing an overload. If the problem persists, it may be best to contact a professional electrician for assistance.

## What are the dangers of electricity problems?

Electricity problems can pose serious safety hazards, such as electrical fires, electric shocks, and damage to appliances and electronics. It is important to address any electricity problems promptly to prevent these dangers.

## How can I prevent electricity problems in my home?

To prevent electricity problems, it is important to have your electrical system regularly inspected by a professional. Avoid overloading circuits and make sure all wiring and outlets are in good condition. It is also important to use surge protectors for valuable electronics and appliances.

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