A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R' is:
(a) 1/25
(b) 1/5
(c) 5
(d) 25
Correct answer: (d) 25
Explanation:
We know that,
R ∝ l
Initial resistance of the wire = RΩ
When wire is cut into 5 pieces, its resistance will become \(\frac{1}{5}\)th
Resistance of each piece of wire after cut into 5 pieces = \(\frac{R}{5}\)Ω
Now, these wires are connected in parallel
We know that, in parallel connection,
equivalent resistance is given by
\(\frac{1}{R'}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\) + \(\frac{1}{R_4}\) + \(\frac{1}{R_5}\)
\(\frac{1}{R'}\) = \(\frac{1}{\frac{R}{5}}\) + \(\frac{1}{\frac{R}{5}}\) + \(\frac{1}{\frac{R}{5}}\) + \(\frac{1}{\frac{R}{5}}\) + \(\frac{1}{\frac{R}{5}}\)
\(\frac{1}{R'}\) = \(\frac{5}{R}\) + \(\frac{5}{R}\) + \(\frac{5}{R}\) + \(\frac{5}{R}\) + \(\frac{5}{R}\)
\(\frac{1}{R'}\) = \(\frac{25}{R}\) =
R' = \(\frac{R}{25}\)
Finding \(\frac{R}{R'}\)
\(\frac{R}{R'}\) = \(\frac{R}{\frac{R}{25}}\)
\(\frac{R}{R'}\) = \(\frac{25 R}{R}\)
\(\frac{R}{R'}\) = 25