Question

Two wires that are made up of same material have length in ratio 4:3 and area of cross-section in ratio 5:4. When wires are connected in parallel across a voltage source, the ratio of current flowing through them is​

Answer 1

Explanation:

⇒R

1

:R

2

:R

3

=

A

1

L

1

:

A

2

L

2

:

A

3

L

3

=

A

1

1

:

2A

1

L

2

:

2

A

1

L

3

=1:

2

2

:

2

1

3

R

1

:R

2

:R

3

=1:1:6

As they are in parallel Iα

R

1

⇒I

1

:I

2

:I

3

=

R

1

1

:

R

2

1

:

R

3

1

=1:1:

6

1

=6:6:1

Power=VI⇒PαI

⇒P

1

:P

2

:P

3

=I

1

:I

2

:I

3

=6:6:1

Answer 2

Given:

Two wires that are made up of same material have length in ratio 4:3 and area of cross-section in ratio 5:4.

To find:

When wires are connected in parallel across a voltage source, the ratio of current flowing through them is ?

Calculation:

Since the wires have same material, they will have same resistivity.

$\therefore \: \dfrac{R_{1}}{R_{2}} = \dfrac{ \rho( \frac{l_{1}}{a_{1}} )}{ \rho( \frac{l_{2}}{a_{2}} )}$

$\implies \: \dfrac{R_{1}}{R_{2}} = \dfrac{l_{1}}{l_{2}} \times \dfrac{a_{2}}{a_{1}}$

$\implies \: \dfrac{R_{1}}{R_{2}} = \dfrac{4}{3} \times \dfrac{4}{5}$

$\implies \: \dfrac{R_{1}}{R_{2}} = \dfrac{16}{15}$

Taking R as constant of proportionality :

$\therefore \: R_{1} = 16R \: and \:R_{2} = 15R$

In parallel connection, the potential drop along each resistance remains same.

So, current through R1 :

$\therefore \: i_{1} = \dfrac{V}{16R}$

So, current through R2 :

$\therefore \: i_{2} = \dfrac{V}{15R}$

So, required ratio :

$\boxed{ \bf\therefore \: i_{1} : i_{2} = 15 : 16}$