two wires of the same metal have the same length but their cross sections are in the ratio 3:1. They are joined in series. The resistance of the thicker wire is 10 ohm. Find the total resistance of the combination.
Answers
Answered by
138
resistence of wire A =10 ohm.
therefore resistence of thinner wire=3*10=30 ohm. (R=pl/A)
total R = 10+30=40 ohm..(since they are in series)
therefore resistence of thinner wire=3*10=30 ohm. (R=pl/A)
total R = 10+30=40 ohm..(since they are in series)
Answered by
59
resistance of thicker wire = 10 Π
as cross section is inversely proportional to resistance
resistance of smaller wire be r and bigger be R
R=¶l/3a
r=¶l/a
so resistance of bigger wire = r/3
R=r/3
R=10Π
r=10*3=30Π
total resistance =30Π+10Π=40Π
answer =40Π
as cross section is inversely proportional to resistance
resistance of smaller wire be r and bigger be R
R=¶l/3a
r=¶l/a
so resistance of bigger wire = r/3
R=r/3
R=10Π
r=10*3=30Π
total resistance =30Π+10Π=40Π
answer =40Π
Similar questions