Question

The length of four conducting wires are in the ratio 1:2:3:4. All wires are of the same material and same radius. If they are connected in parallel with a battery. Then the ratio of currents in these wires will be
A. $12:6:4:3$
B. $6:3:2:1$
C. $4:3:2:1$
D. $1:2:3:4$

Answer 1

According to the current divider rule, in parallel connection the voltage is the same at the junction point and in series connection the current is the same in the wire.In this question the nature of the material is the same for all the wires. The resistivity of wire represents the nature of the material. So, the resistivity of all the wires is the same.

Complete step by step solution:
There are four wires having same material i.e.,
${{\rho }_{1}}={{\rho }_{2}}={{\rho }_{3}}={{\rho }_{4}}$
All the wires are of same radius i.e. ${{r}_{}}={{r}_{}}={{r}_{}}={{r}_{4}}$
The length of the four conducting wire i.e. ${{L}_{}}:{{L}_{}}:{{L}_{}}:{{L}_{4}}=12:6:4:3$
All the wire is connected in parallel with the battery..

To find out the ratio of the currents in wires
Let the current across each wire be $I,\text{ }I,I\text{ }\And \text{ }I$ respectively.
According to the ohm’s laws
$I=\dfrac{V}{R}$
The wires are connected in parallel with a battery, therefore, according to the current divider rule as the voltage is same in parallel connection, so we can write,

$$I:\text{ }I:I\text{ : }I=\dfrac{V}{{{R}_{1}}}:\dfrac{V}{{{R}_{2}}}:\dfrac{V}{{{R}_{3}}}:\dfrac{V}{{{R}_{4}}} \\\ \Rightarrow I:\text{ }I:I\text{ : }I=\dfrac{1}{{{R}_{1}}}:\dfrac{1}{{{R}_{2}}}:\dfrac{1}{{{R}_{3}}}:\dfrac{1}{{{R}_{4}}} \\\$$

Whereas
$R=\dfrac{\rho L}{A}$
R is the Resistance of the wire, L is the length of the wire, A is the area of the cross-section
$\rho$ is the resistivity of the wire. The wire is circular cross-section, so the area of the circular cross-section of the wire is $\pi {{r}^{2}}$.

$$\dfrac{1}{{{R}_{1}}}:\dfrac{1}{{{R}_{2}}}:\dfrac{1}{{{R}_{3}}}:\dfrac{1}{{{R}_{4}}} \\\ \Rightarrow \dfrac{{{A}_{1}}}{{{\rho }_{1}}{{L}_{1}}}:\dfrac{{{A}_{2}}}{{{\rho }_{2}}{{L}_{2}}}:\dfrac{{{A}_{3}}}{{{\rho }_{3}}{{L}_{3}}}:\dfrac{{{A}_{4}}}{{{\rho }_{4}}{{L}_{4}}} \\\$$

(All the wire is same radius so the area of the cross-section will be same)
$\dfrac{1}{{{\rho }_{1}}{{L}_{1}}}:\dfrac{1}{{{\rho }_{2}}{{L}_{2}}}:\dfrac{1}{{{\rho }_{3}}{{L}_{3}}}:\dfrac{1}{{{\rho }_{4}}{{L}_{4}}}$ $({{A}_{1}}={{A}_{2}}={{A}_{3}}={{A}_{4}})$

All the wire is same material $({{\rho }_{1}}={{\rho }_{2}}={{\rho }_{3}}={{\rho }_{4}})$
$\dfrac{1}{{{L}_{1}}}:\dfrac{1}{L{}_{2}}:\dfrac{1}{{{L}_{3}}}:\dfrac{1}{{{L}_{4}}}$
$\Rightarrow\dfrac{1}{1}:\dfrac{1}{2}:\dfrac{1}{3}:\dfrac{1}{4}$
Multiply with 12

$$\dfrac{1}{1}\times 12:\dfrac{1}{2}\times 12:\dfrac{1}{3}\times 12:\dfrac{1}{4}\times 12 \\\ \Rightarrow 12:6:4:3$$

Thus, the ratio of currents in the four conducting wires, $I\text{ }:\text{ }I\text{ }:\text{ }I\text{ }:\text{ }I\text{ =}~12\text{ }:\text{ }6\text{ }:\text{ }4\text{ }:\text{ }3$

So, the correct answer is option A.

Note: While solving this problem, most of the students have a common confusion that resistivity, area of cross section and the length of the wire is related to resistance. They seem confused at the point that how does current come in the calculation of ratio. But it is important to remember that as the potential difference is the same across all the wires, current is inversely proportional to the resistance.