(A) The total resistance of the wire is $R = 20 \Omega$. It is divided into $10$ equal parts,so the resistance of each part is $r = \frac{20 \Omega}{1

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(A) The total resistance of the wire is $R = 20 \Omega$. It is divided into $10$ equal parts,so the resistance of each part is $r = \frac{20 \Omega}{10} = 2 \Omega$.Two parts are connected in parallel. The equivalent resistance of one such parallel pair is $R_p = \frac{r \times r}{r + r} = \frac{2 \times 2}{2 + 2} = 1 \Omega$.Since there were $10$ parts in total,and we used them in pairs,we have $10 / 2 = 5$ such parallel pairs.These $5$ pairs are then connected in series. The total equivalent resistance is $R_{eq} = 5 \times R_p = 5 \times 1 \Omega = 5 \Omega$.

Class 12 Physics Current Electricity Equivalent Resistance - Series and Parallel , Circuit

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$A$ wire of resistance $20 \Omega$ is divided into $10$ equal parts. $A$ combination of two parts are connected in parallel and so on. Now,the resulting pairs of parallel combinations are connected in series. The equivalent resistance of the final combination is . . . . . . $\Omega$.

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$A$ wire of resistance $20 \Omega$ is divided into $10$ equal parts. $A$ combination of two parts are connected in parallel and so on. Now,the resulting pairs of parallel combinations are connected in series. The equivalent resistance of the final combination is . . . . . . $\Omega$.

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$A$ technician has only two resistance coils. By using them singly, in series or in parallel, he is able to obtain the resistances $3, 4, 12$ and $16 \, \Omega$. The resistances of the two coils are ........... $\Omega$.

$A$ wire of length $l$ and resistance $100 \Omega$ is divided into $10$ equal parts. The first $5$ parts are connected in series while the next $5$ parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is: (in $Omega$)

$A$ uniform conductor of resistance $R$ is cut into $20$ equal pieces. Half of them are joined in series and the remaining half of them are connected in parallel. If the two combinations are joined in series,the effective resistance of all the pieces is

$A$ wire is connected to a battery between the points $M$ and $N$ as shown in figure $(1)$. The same wire is bent in the form of a square and then connected to the battery between the points $M$ and $N$ as shown in figure $(2)$. Which of the following quantities increases?

Three resistors are connected to form the sides of a triangle $ABC$. The resistances of the sides $AB$,$BC$,and $CA$ are $40\,\Omega$,$60\,\Omega$,and $100\,\Omega$ respectively. The effective resistance between the points $A$ and $B$ in $\Omega$ will be:

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