A length of wire is cut into five equal pieces.?
The five pieces are then connected in parallel,
with the resulting resistance being 78.0 Ω.
What was the resistance of the original
length of wire before it was cut up?
Answer in units of Ω
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Verified answer
390 Ohms.
1/R total = 1/R1 + 1/R2 +1/R3 +1/R4 +1/R5, where R1=R2=R3=R4=R5
1/78 = 5(1/R)
0.0128 = 5(1/R)
0.0026 = 1/R
1/0.0026 = R
R = 390 Ohms
Hope this helps,
Newton1Law
Equivalent of N equal resistors of value R each = R ÷ N
So 78 = R ÷ 5
R = 78 x 5 = 390 ohms each
Total resistance of the original length of wire = 1950 ohms
The resistance of copper cord is so low that the joints will probable have extra resistance than the cord. So it relatively is attainable that your little venture has bigger resistance than a single piece of cord. yet as quickly as we someway stay away from that situation then the completed resistance of N products in parallel is a million/N cases the resistance of one of the products. (And the resistance of each bit is a million/N cases the unique cord.)