Question
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
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.
Related Questions
Two copper wires have equal cross-sectional areas and lengths of 2.4m and 1.3m, respectively.
What...
Two wires have the same length and the same resistance but are made up of different materials. The r...
Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.96 cm^2...