
SOLUTION :
Resistance is directly proportional to length of wire and inversely proportional to cross sectional area of wire. The constant of proportionality (p) is termed as resistivity of wire with respect length and area.
So,
R1
= p1 * l1 / A1
= 364 * 10^(-6) * 4.1 / (pi * (1*10^(-3))^2)
= 475.046 Ω
R2
= p2 * l2 / A2
= 261 * 10^(-6) * 2.6 / (pi * (8*10^(-3))^2)
= 3.375 Ω
R1 an R2 are in series.
So,
Total resistance of the joint system, R
= R1 + R2
= 475.046 + 3.375
= 478.421
= 478.4 Ω (ANSWER)
Consider the two cylindrical wires of lengths ly=4.1m and lz=2.6m are merged back to back as...
Consider the two cylindrical wires of lengths 41=3.8m and 1=2.2m are merged back to back as shown in the figure below to form a single wire. The wires have the radii of r1=1mm and r2=9mm and the current runs along the lengths of the wires. If the thin and the thick wires resistivities are $1=108x10-60.m and P2=337x10 60.m respectively, determine the resistance of the wire. Please express your answer in units of using one decimal places. l2 li 11 12
Consider the two cylindrical wires of lengths &q=4.8m and 62=3.7m are merged back to back as shown in the figure below to form a single wire. The wires have the radii of r1=1mm and rz=8mm and the current runs along the lengths of the wires. If the thin and the thick wires resistivities are p1=171x10-60.m and p2=391x10-60.m respectively, determine the resistance of the wire. Please express your answer in units of using one decimal places. 12 li 11 72
Consider the two cylindrical wires of lengths l1 =3.8m and l2=2m are merged back to back as shown in the figure below to form a single wire. The wires have the radii of r1=3mm and r2= 8mm and the current runs along the lengths of the wires. If the thin and the thick wires resistivities are p1=152x10-6Ω.m and p2=201x10-6Ω.m respectively, determine the resistance of the wire. Please express your answer in units of Ω using one decimal places.