Question

Two wires of the same material have length $3\,cm$ and $5\,cm$ and radii $1\,mm$ and $3\,mm$ respectively. They are connected in series across a battery of $16\,V$. The p.d. across the shorter wire is:
A) $2.5\,V$
B) $6.5\,V$
C) $1.5\,V$
D) $13.5\,V$

Answer 1

If the material is made of the same material, then its resistivity is also the same. Hence compare the resistivity of the two wires, from that obtain the value of the resistance. Substitute this in the ohm’s law given , and find the current through the battery and the voltage.

Formula used:
(1) The formula of the resistivity is given by
$\rho = \dfrac{{RA}}{L}$
Where $\rho$ is the resistivity of the material of the wire, $R$ is the resistance of the wire, $A$ is the area of the wire and $L$ is the length of the wire.
(2) ohm’s law is given by
$V = IR$
Where $V$is the potential developed across the wires and $I$ is the current passing through the circuit.

Complete step by step solution:
The length of the wire, ${l_1} = 3\,cm$
The length of the other wire, ${l_2} = 5\,cm$
Radius of the first wire, ${r_1} = 1\,mm$
Radius of the second wire, ${r_2} = 3\,mm$
The emf of the batter, $e = 16\,V$
It is given that the material is the same, and hence the resistivity of the material is also the same. By equating the resistivity of the two wires.
$\Rightarrow$ $\dfrac{{{R_1}{A_1}}}{{{L_1}}} = \dfrac{{{R_2}{A_2}}}{{{L_2}}}$
By rearranging the above equation, we get
$\Rightarrow$ $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{A_2}{L_1}}}{{{A_1}{L_2}}}$
By calculating the area of the wire from substituting the radius in the formula of the area,
$\Rightarrow$ $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{2.25\pi \times 60}}{{2.25\pi \times 100}}$
By simplifying the above equation, we get
$\Rightarrow$ $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{27}}{5}$
Hence the value of the ${R_1}$ is $27$ and the value of the ${R_2}$ is $5$ .
Both the resistors are in series, hence the total resistance is $27 + 5 = 32$
Let us calculate the current from the battery by the formula of the ohm’s law.
$i = \dfrac{V}{R}$
$\Rightarrow$ $i = \dfrac{{16}}{{32}} = 0.5\,A$
The same current from the battery moves through the wires. Hence
$V = iR$
Substitute the values,
$\Rightarrow$ $V = 0.5 \times 27 = 13.5\,V$
Hence the potential difference across the short wire is $13.5\,V$.

Thus the option (D) is correct.

Note: If the circuit is in series connection, then the current flowing through all these will be the same and the total resistance in the circuit is the sum of the resistance in each branch of the circuit. But in parallel circuits, the voltage is similar.