A 4 ohm resistance wire is doubled on it. Calculate the new resistance of the wire.
Answers
Explanation:
We consider wire is doubled on it means to fold the length of wire. It means its length will get half and area of cross section will get doubled.
∴ Let the resistance of the wire originally 'R' of length 'L' and area of cross-section 'A' with resistivity of material is 'p',
Then
R = p l/A = 4Ω
Now, for new arrangement,
p' = p
l' = l / 2
A' = 2A
Thus,
R' = p' l'/A' = p (l /2) ÷ 2A = !/4 (p l / A) = 1/4R = 1/4 × 4 = 1Ω.
Or
When the wire is doubled the length will be halved and area will be double or twice
When the wire is doubled the length will be halved and area will be double or twice R = PL/A
When the wire is doubled the length will be halved and area will be double or twice R = PL/A R1 = P(L/2) / 2A Where R1 is the new resistance
When the wire is doubled the length will be halved and area will be double or twice R = PL/A R1 = P(L/2) / 2A Where R1 is the new resistance So R1/R = {P(L/2) / 2A} / {PL/A}
When the wire is doubled the length will be halved and area will be double or twice R = PL/A R1 = P(L/2) / 2A Where R1 is the new resistance So R1/R = {P(L/2) / 2A} / {PL/A} When we solve this we get R1/R = ¼
When the wire is doubled the length will be halved and area will be double or twice R = PL/A R1 = P(L/2) / 2A Where R1 is the new resistance So R1/R = {P(L/2) / 2A} / {PL/A} When we solve this we get R1/R = ¼ => R1 = R/4 = 4/4 = 1 OHM