Four copper wires of equal length L are connected to a 100V source. Their cross-sectional area are 1cm^2, 2cm^2, 3cm^2, and 4cm^2. Find the voltage across the 3cm^2 wire. Leave your answer in terms of L, it should cancel out.
- So I tried using and equating various relevant equations together such as E = pJ, ΔV = -EL, ΔV = pLI/A, and R = pL/A but I keep ending up with more than one unknown or an unknown that is not what I'm looking for (which is voltage across the 3cm^2 wire). Can someone please help me with this? Thanks you.
p = 1.68*10^-8 Ohm m = 1.68*10^-6 Ohm cm.
R1 = pL/A = (1.68*10^-6)L/1cm^2 =
(1.68*10^-6)L Ohm.
R2 = (1.68*10^-6)L/2cm^2 = (0.84*10^-6)L Ohm.
R3 = (0.56*10^-6)L Ohm.
R4 = (0.42*10^-6)L Ohms.
R = R1 + R2 + R3 + R4.
R = (1.68L+0.84L+0.56L+0.42L)10^-6 = (3.5*10^-6)L Ohms.
I = E/R = 100/(3.5*10^-6)L = (28.57*10^6)/L Amps.
V3 - I*R3 = 28.57*10^6/L * (0.56*10^-6)L = 16 Volts.
To find the voltage across the 3cm^2 wire, we can use the concept of resistivity and the equation:
R = pL/A
Where R is the resistance of the wire, p is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire.
In this case, we have four copper wires of equal length L, with cross-sectional areas of 1cm^2, 2cm^2, 3cm^2, and 4cm^2.
Let's assume that the resistivity of copper is p, and the resistances of the wires are R1, R2, R3, and R4. We can express the resistances as:
R1 = pL/1cm^2
R2 = pL/2cm^2
R3 = pL/3cm^2
R4 = pL/4cm^2
Given that the wires are connected to a 100V source, and considering the equivalent resistance of the wires in series:
V = I * Req
Where V is the applied voltage (100V), I is the current flowing through the circuit, and Req is the equivalent resistance of the wires.
Since the wires are connected in series, the current flowing through each wire is the same. So, we can write:
I = V / (R1 + R2 + R3 + R4)
Substituting the expressions for R1, R2, R3, and R4:
I = V / (pL/1cm^2 + pL/2cm^2 + pL/3cm^2 + pL/4cm^2)
I = V / (pL(1/1cm^2 + 1/2cm^2 + 1/3cm^2 + 1/4cm^2))
Now, we can find the resistance of the 3cm^2 wire by substituting the values into the equation for resistance:
R3 = pL/3cm^2
Finally, let's rewrite the expression for the current (I) using resistance R3:
I = V / (R1 + R2 + R3 + R4)
I = V / (pL/1cm^2 + pL/2cm^2 + pL/3cm^2 + pL/4cm^2)
I = V / (pL(1/1cm^2 + 1/2cm^2 + 1/3cm^2 + 1/4cm^2))
I = V / (pL(60/(12cm^2)))
I = V / (5pL/cm^2)
Now, we can find the voltage across the 3cm^2 wire (ΔV3) using Ohm's Law:
ΔV3 = I * R3
ΔV3 = (V / (5pL/cm^2)) * (pL/3cm^2)
ΔV3 = V * (1 / (5cm^2 * 3cm^2))
ΔV3 = V / 15cm^4
Therefore, the voltage across the 3cm^2 wire is V / 15cm^4 (in terms of L, it will cancel out).
To find the voltage across the 3cm^2 wire, we can make use of the fact that the current is the same in all the wires (assuming they are connected in parallel).
Let's assume the resistivity of copper (p) is constant. Since the cross-sectional areas (A) and lengths (L) of the wires are given for each wire, we can calculate the resistance (R) of each wire using the formula:
R = pL/A
Now, let's consider the 3cm^2 wire. We can represent its resistance as R3:
R3 = pL/A3
Given that the voltage across the entire circuit is 100V, and all the wires are connected in parallel, the voltage across each wire will be the same. So, the voltage across the 3cm^2 wire can be represented as V3.
Now, we can use Ohm's Law to relate the resistance and voltage:
V = IR
Since the current (I) is the same for all the wires, we can substitute I with V/R:
V = V3
V = (V3)pL/A3
We can now solve for V3 by isolating it in this equation. To do that, we can rearrange the formula as follows:
(V3)pL/A3 = 100
Now, divide both sides of the equation by pL/A3:
V3 = 100A3/(pL)
Thus, the voltage across the 3cm^2 wire is given by:
V3 = 100A3/(pL)
Notice that the length (L) will get canceled out when we substitute the appropriate values for A3 and p.