A wire of length l and resistance 100 Omega i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The resistance of the original wire is $R = 100 \Omega$.When the wire is divided into $10$ equal parts,the resistance of each part is $r = \frac{R

Vedclass · Accepted Answer

$52$

Vedclass · Answer

(A) The resistance of the original wire is $R = 100 \Omega$.When the wire is divided into $10$ equal parts,the resistance of each part is $r = \frac{R}{10} = \frac{100 \Omega}{10} = 10 \Omega$.For the first $5$ parts connected in series,the equivalent resistance is $R_S = 5 	imes r = 5 	imes 10 \Omega = 50 \Omega$.For the next $5$ parts connected in parallel,the equivalent resistance $R_P$ is given by $\frac{1}{R_P} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} = \frac{5}{r}$.Thus,$R_P = \frac{r}{5} = \frac{10 \Omega}{5} = 2 \Omega$.Since these two combinations are connected in series,the final equivalent resistance is $R_{eq} = R_S + R_P = 50 \Omega + 2 \Omega = 52 \Omega$.

$A$ 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 $Omega$)

Similar Questions

Resistance $n$,each of $r \ \Omega$,when connected in parallel give an equivalent resistance of $R \ \Omega$. If these resistances were connected in series,the combination would have a resistance in $\Omega$,equal to

You are given $n$ resistors each of resistance $r$. They are first connected to get the minimum possible resistance. In the second case,they are again connected differently to get the maximum possible resistance. Calculate the ratio between minimum and maximum values of resistance so obtained.

The effective resistance across $AB$ in the network shown is .......... $\Omega$.

In the figure shown,what is the current (in Ampere) drawn from the battery? You are given $R_1 = 15\,\Omega$,$R_2 = 10\,\Omega$,$R_3 = 20\,\Omega$,$R_4 = 5\,\Omega$,$R_5 = 25\,\Omega$,$R_6 = 30\,\Omega$,and $E = 15\,V$.

$10$ resistors,each of resistance $R$,are connected in series to a battery of $emf$ $E$ and negligible internal resistance. When those are connected in parallel to the same battery,the current is increased $n$ times. The value of $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module