Question

Two wires of the same material have length 6cm and 10cm and radii 0.5mm and 1.5 mm respectively They are connected in series across a battery of 16V The potential difference across the shorter wire is

Answer 1

L1 = 6cm = 60mm
L2 = 10cm = 100mm
r1 = 0.5mm
r2 = 1.5mm

Material is same so resistivity would also be same.

A1R1/L1 = A2R2/L2

A1 = pi × 0.5 × 0.5 = 0.25pi

A2 = 2.25pi

R1/R2 = A2L1 / A1L2

R1/R2 = 2.25pi × 60 / 0.25pi × 100

R1/R2 = 225 × 60 / 25 × 100

R1/R2 = 5.4 = 54/10 = 27/ 5

Hence R1 = 27 and R2 = 5

In series total resistance is given by R1 +R2
27 + 5 = 32

i = V/R

i = 16 / 32 = 1/2 = 0.5A

Now p.d across 6cm wire is

V = iR

V = 0.5 × 27

V = 13.5V

Thanks

Answer 2

The potential difference across the shorter wire is 13.5 V

Given,

length of first wire, L1 = 6 cm = 60 mm

length of second wire, L2 = 10 cm = 100 mm

radius of first wire, r1 = 0.5 mm

radius of second wire, r2 = 1.5 mm

total potential difference, V = 16V

To Find,

potential difference across the shorter wire, V1 = ?

Solution,

The resistance of a wire is given by the following formula:

R = ρ × L ÷ A

We have been given two wires 1 and 2, of same material.

Therefore, the resistivity of the two wires, ρ1 = ρ2 = ρ (since the wires are of the same materials).

Therefore, the ratio of the resistances of the two wires can be given by:

$\implies \frac{R1}{R2} = \frac{\rho1 \times L1}{A1} \times \frac{A2}{\rho2 \times L2}\\\\\implies \frac{R1}{R2} = \frac{\rho \times L1}{\pi \times {r1}^2} \times \frac{\pi \times {r2}^2}{\rho \times L2}\\\\\implies \frac{R1}{R2} = \frac{L1}{{r1}^2} \times \frac{{r2}^2}{L2}\\\\\implies \frac{R1}{R2} = \frac{60}{100} \times \frac{2.25}{0.25}\\\\\implies \frac{R1}{R2} = \frac{3}{5} \times 9\\\\\implies \frac{R1}{R2} = \frac{27}{5}$

Therefore, R1 = 27p and R2 = 5p, where p is a constant.

Now, since the wires are connected in series, the total current will be equal to the current flowing through each of the wires.

$\implies I1 = I2 = I\\ \\\implies V1 \times R1 = V \times R$

Here, R is the total resistance

R = R1 + R2, since R1 and R2 are in series

$\implies V1 \times R1 = V \times (R1+R2)\\\\\implies V1 = V \times \frac{R1}{R1+R2}\\\\\implies V1 = 16 \times \frac{27p}{27p+5p}\\\\\implies V1 = 16 \times \frac{27p}{32p}\\\\\implies V1 = \frac{27}{2}\\\\\implies V1 = 13.5 V$

Therefore, the potential difference across the shorter wire is 13.5 V

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