If a wire of resistance 20 Ω is bent to form a closed square, the resistance is ___

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If a wire of resistance $20 Ω$ is bent to form a closed square, the resistance is ____ $Ω$ across a diagonal of the square.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

If a wire of resistance

20 Ω

is bent to form a closed square, the resistance is

5 Ω

across a diagonal of the square.
It is given that the resistance of the wire is

20 Ω

.
The number of sides in a square is

4 .

Therefore, the resistance of a side of the square is

\Rightarrow R = \frac{20}{4}

\Rightarrow R = 5 Ω

The resistance of a single side is

5 Ω

Now, side AB and BC are in a series combination. Also, side AD and DC are in series combination. We know that,
the equivalent resistance of resistances connected in a series combination is given by

R_{s} = R_{1} + R_{2}

Therefore, their equivalent resistance Rs would be

\Rightarrow R_{s} = 5 Ω + 5 Ω

\Rightarrow R_{s} = 10 Ω

Now, both 10

Ω

resistances are in parallel.
Also, the net resistance in parallel combination is given by

\frac{1}{R_{s}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}

Therefore, the equivalent resistance across a diagonal is

\Rightarrow \frac{1}{R_{p}} = \frac{1}{10} + \frac{1}{10}

Ω

\Rightarrow \frac{1}{R_{p}} = \frac{2}{10} Ω

\Rightarrow \frac{1}{R_{p}} = \frac{1}{5} Ω

\Rightarrow R_{p} = 5 Ω

Hence, the equivalent resistance across a diagonal of the square is

5 Ω

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th