JEE Main 2023 — Current Electricity Question with Solution

Our approach to this problem uses the fact that the voltage across the conductor remains constant as it is connected to the same source. By analyzing the relation between the currents and resistances at different temperatures, you can find the current flowing through the conductor at $$50^{\circ }\mathrm{C}$$.

First, you establish a relationship between the currents and resistances at $$0^{\circ }\mathrm{C}$$ and $$100^{\circ }\mathrm{C}$$:

$$i_0{R}_0=i_{100}{R}_{100}$$

Plugging in the given values for $$i_0$$ and $$i_{100}$$:

$$2R_0=1.2R_0(1+100\alpha )\cdots (1)$$

From this equation, you find the value of $$\alpha$$:

$$1+100\alpha =\frac{5}{3}\Rightarrow 100\alpha =\frac{2}{3}\Rightarrow 50\alpha =\frac{1}{3}$$

Now, you need to find the current $$i_{50}$$ at $$50^{\circ }\mathrm{C}$$. To do this, you calculate the resistance $$R_{50}$$ using the found value of $$\alpha$$:

$$R_{50}=R_0(1+50\alpha )=R_0(1+\frac{1}{3})$$

Using the fact that the voltage across the conductor remains constant, you can find the current $$i_{50}$$:

$$i_{50}=\frac{i_0{R}_0}{R_{50}}=\frac{2\times R_0}{R_0(1+\frac{1}{3})}=\frac{2}{1+\frac{1}{3}}=1.5\mathrm{A}$$

Thus, the current flowing through the conductor at $$50^{\circ }\mathrm{C}$$ is $$15\times 10^2\mathrm{mA}$$