Question

One end of a nichrome wire of length $2L$ and cross-sectional area $A$ is attached to an end of another nichrome wire of length $L$ and cross-sectional area $2A$. If the free end of the longer wire is at an electric potential of $8.0\text{Volts}$, and the free end of the shorter wire is at an electric potential of $1.0\text{Volts}$, the potential at the junction of the two wires is equal to

• A. $2.4\text{V}$
• B. $3.2\text{V}$
• C. $4.5\text{V}$
• D. $5.6\text{V}$

Answer 1

The correct option is A $2.4\text{V}$

As we know that, resistance of a wire can be defined as

$R=\frac{\rho l}{A}$


So, resistance of longer wire $R$ and shorter wire $R^{\prime }$ is given by

$R=\frac{\rho \left(2L\right)}{A}=\frac{2\rho L}{A}$

$R^{\prime }=\frac{\rho \left(L\right)}{2A}=\frac{\rho L}{2A}=\frac{R}{4}$

Here $\rho$ will be same for both wires because both wire are made of nichrome and same current will flow through the two wires due to series combination.

Let the potential at their junction be $x$ then,


Apply KCL at junction $X$

$i_1=i_2$

$\Rightarrow \frac{8-x}{R}=\frac{x-1}{R/4}$

$\Rightarrow 8-x=4x-4$

$\therefore x=\frac{12}{5}\text{V}=2.4\text{V}$

Hence, option $\left(a\right)$ is correct.

Key concept- When resistances are connected end to end in series then same current flows through them.