Asked by Kid
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.
- 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.
Answers
Answered by
Henry
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.
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.
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