You visited us 1 times! Enjoying our articles? Unlock Full Access!

Resistance

A conductor w...

Question

A conductor with rectangular cross-section has dimensions $(a \times 2 a \times 4 a)$ as shown in figure. Resistance across AB is $R_{1}$ , across CD is $R_{2}$ and across EF is $R_{3}$ . Then which of the following is true?

R_{1} = R_{2} = R_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{1}

R_{2}

R_{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

R_{2}

R_{3}

R_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{1}

R_{3}

R_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $R_{1}$ > $R_{2}$ > $R_{3}$

Resistance $R = ρ \frac{l}{A}$
From the figure the dimensions are $a$ , $2 a$ and $4 a$ , i.e.
Distance between faces A and B = $4 a$
Distance between faces C and D= $2 a$
Distance between faces E and F= $a$

As $R \propto l$ , AB having the highest length will have the most resistance followed by CD and EF being of the lowest length will have the least resistance. So, the order is $R_{1} > R_{2} > R_{3}$ .

Suggest Corrections

Similar questions

L - C - R

ω

R_{1}, R_{2} a n d R_{3}

△ A B C

a < b < c

r

r_{1}, r_{2}, r_{3}

A, B, C

r < r_{1} < r_{2} < r_{3}

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

Join BYJU'S Learning Program