A uniform conductor of resistance R is cut in | vedclass

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

Question

(C) Resistance of each piece $= \frac{R}{20} \Omega$. Equivalent resistance of $10$ such pieces connected in series is $R_1 = 10 	imes \frac{R}{20} =

Vedclass · Accepted Answer

$\frac{101 R}{200}$

Vedclass · Answer

(C) Resistance of each piece $= \frac{R}{20} \Omega$. Equivalent resistance of $10$ such pieces connected in series is $R_1 = 10 	imes \frac{R}{20} = \frac{R}{2}$. Similarly,equivalent resistance of $10$ such pieces connected in parallel is $\frac{1}{R_2} = \frac{1}{R/20} + \dots (10 	ext{ times}) = \frac{10}{R/20} = \frac{200}{R}$,so $R_2 = \frac{R}{200}$. Now,$R_1$ and $R_2$ are connected in series,so the total resistance is $R_{eq} = R_1 + R_2 = \frac{R}{2} + \frac{R}{200} = \frac{100R + R}{200} = \frac{101R}{200}$.

$A$ 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

Similar Questions

For the reaction $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,$K_c = 1.8 \times 10^{-6}$ at $185 \ ^oC$. At $185 \ ^oC$,the $K_c$ for the reaction $NO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons NO_{2(g)}$ is:

The equation of a straight line,perpendicular to $3x - 4y = 6$ and forming a triangle of area $6$ sq. units with the coordinate axes,is

The figure shows a system of logic gates. From the study of the truth table,it can be found that to produce a high output $(1)$ at $R$,we must have

Which of the following statements is incorrect?

The value of '$a$' for which the three given planes
$P_1: (a+1)x - y - z = a$
$P_2: x - (a+1)y - z = a$
$P_3: x - y - (a+1)z = a$
have no point in common,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module