Question

Two metallic wires A & B are connected in series wire A has length like & radius r & wire B has length 2l & radius 2r find the ratio of the total resistance of series combination & the resistance of wire A , if both are of some materials

Question 6

Answer 1

we know that
R=pt/a
according to the given data
=length of wire A=A
=length of wire B=2B
=radius of wire A=R
=radius of wire B=2R
let resists of wire be Ra
=Ra=pt/πr^2
resists of wire B(Rb)
=p(2a)/π(2r)
equalent resists of series
Ra+Rb
Ra+1/2Rb
3/2Rb
∴ratio
3/2Ra/Ra
3/2

Answer 2

Answer:

Let the resistivity of both the wires be 'ρ'.

The resistance of the wire A is given as R

A

=

πr

2

ρl

.

The resistance of the wire B is given as R

B

=

π2r

2

ρ2l

=

2πr

2

ρl

.

Now when the resistance are connected in seires is R=R

A

+R

B

. That is, R=

2

3

(

πr

2

ρl

).

So, the required ratio is R:R

A

=

2

3

(

πr

2

ρl

):

πr

2

ql

.

Therefore, R:R

A

=3:2.

Hence, the ratio of the resistances in series and only wire A is 3/2.