Question

2. A metallic wire of resistance R is cut into ten equal parts of equal length. Two pieces each are joined in series and then five such combinations are joined in parallel .What will be the effective resistance in such case

Answer 1

When a metallic wire of resistance R is cut into 10 equal length. Then the new resistance becomes R' = R/10.

Given that, two pieces each are joined in series and then five such combinations are joined in parallel.

For series:

Rs = R1 + R2

As they are of same resistance also wire is same. So,

Rs = R' + R'

Rs = 2R'

Rs = 2 × R/10

Rs = R/5 ohm

For parallel:

Before that read the question again, "Two pieces each are joined in series and then five such combinations are joined in parallel."

The value of series combination which is we get is then connected in parallel. So,

1/Rp = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5

1/Rp = 1/R' + 1/R' + 1/R' + 1/R' + 1/R'

1/Rp = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)

1/Rp = 5/R + 5/R + 5/R + 5/R + 5/R

1/Rp = 5/R (1 + 1 + 1 + 1 + 1)

1/Rp = 5/R (5)

1/Rp = 25/R

Rp = R/25 ohm

Req = Rp

Req = R/25 ohm

Answer 2

$\Large\underline{\underline{\sf{\color{orange}{GIVEN}}}}$

☛ A metallic wire of resistance R is cut into ten equal parts of equal length.

☛ Two pieces each are joined in series and then five such combinations are joined in parallel.

$\Large\underline{\underline{\sf{\color{orange}{TO\ FIND}}}}$

✏ The effective resistance in such case.

$\Large\underline{\underline{\sf{\color{orange}{SOLUTION}}}}$

A wire is cut into 10 EQUAL PARTS of EQUAL LENGTH.

The resistance of each part = R/10 Ω.

We know,

$\sf{\bigstar\ For\ series\ connections,}\\\displaystyle{\sf{R_s=R_1+R_2+R_3+...+R_n }}\\\\\sf{\bigstar\ For\ parallel\ connections,}\\\displaystyle{\sf{\frac{1}{R_p} =\frac{1}{R_1} +\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n} }}$

$\rule{190}1$

When 2 pieces are connected in series,

$\sf{R_s=\dfrac{R}{10} +\dfrac{R}{10} }\\\\\sf{\longmapsto R_s=\dfrac{2R}{10} }\\\\\sf{\longmapsto R_s=\dfrac{R}{5} }$

Given, 5 such combinations whose equivalent resistance is R/5 Ω, are joined in parallel. (i.e. 5 such series connections of 2 pieces are connected in parallel).

$\displaystyle{\sf{\frac{1}{R_p} =\frac{1}{R_1} +\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4} +\frac{1}{R_5} }}\\\\\\\displaystyle{\longmapsto \sf{\frac{1}{R_p} =\frac{1}{\frac{R}{5} } +\frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}} +\frac{1}{\frac{R}{5}} }}\\\\\\\displaystyle{\sf{\longmapsto \frac{1}{R_p} =\frac{5}{R} +\frac{5}{R}+\frac{5}{R}+\frac{5}{R} +\frac{5}{R} }}$

$\displaystyle{\sf{\longmapsto \frac{1}{R_p} =\frac{5+5+5+5+5}{R} }}\\\\\\\\\displaystyle{\sf{\longmapsto \frac{1}{R_p} =\frac{25}{R} }}\\\\\\\boxed{\boxed{\displaystyle{\sf{\longmapsto R_p=\frac{R}{25}\ \Omega }}}}$

➥ Hence, the answer is R/25 Ω.