The current flowing through a conductor connected across a source is 2 A and 1.2 A at 0^ C and 100^ C respectively. The current flowing through the conductor at 50^ C will be 10^2 mA.

Given: Current is 2 A at 0^ C and 1.2 A at 100^ C for the same source. Find: Current at 50^ C. For the same source, voltage remains constant, so i0 R0 = i100 R100 Using R100 = R0(1 + 100), 2R0 = 1.2R0(1 + 100) 1 + 100 = 53 100 = 23 Therefore, 50 = 13 Now at 50^ C, i50R50 = i0R0 with R50 = R0(1 + 50) So, i50 = i0R0R50 = 2R0R0(1 + 50) = 21 + 13 = 2 43 = 1.5 A Thus, 1.5 A = 1500 mA = 15 10^2 mA Therefore, the required answer is 15.

The current flowing through a conductor connected across a source… — Solution | JEE Physics

NVAMediumJEE 2023Resistivity & Conductivity

The current flowing through a conductor connected across a source is $2 \, \text{A}$ and $1.2 \, \text{A}$ at $0^\circ\text{C}$ and $100^\circ\text{C}$ respectively. The current flowing through the conductor at $50^\circ\text{C}$ will be _____ $\times 10^2$ $\text{mA}$ .

Answer

Correct answer:15

Step-by-step solution

Standard Method

Given: Current is $2 \, \text{A}$ at $0^\circ\text{C}$ and $1.2 \, \text{A}$ at $100^\circ\text{C}$ for the same source.

Find: Current at $50^\circ\text{C}$ .

For the same source, voltage remains constant, so

i_0 R_0 = i_{100} R_{100}

Using $R_{100} = R_0(1 + 100\alpha)$ ,

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

1 + 100\alpha = \frac{5}{3}

100\alpha = \frac{2}{3}

Therefore,

50\alpha = \frac{1}{3}

Now at $50^\circ\text{C}$ ,

i_{50}R_{50} = i_0R_0

with

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

So,

i_{50} = \frac{i_0R_0}{R_{50}} = \frac{2R_0}{R_0(1 + 50\alpha)} = \frac{2}{1 + \frac{1}{3}} = \frac{2}{\frac{4}{3}} = 1.5 \, \text{A}

Thus,

1.5 \, \text{A} = 1500 \, \text{mA} = 15 \times 10^2 \, \text{mA}

Therefore, the required answer is $15$ .

Common mistakes

Using current directly proportional to temperature is incorrect because the source voltage is the same, so current changes inversely with resistance. First relate resistance with temperature, then use $I = \frac{V}{R}$ .
Substituting $100\alpha$ in place of $50\alpha$ at $50^\circ\text{C}$ is wrong. Since temperature rise is halved, use $50\alpha = \frac{1}{3}$ , not $\frac{2}{3}$ .
Writing the final numerical answer as $1500$ instead of the asked form can be incorrect here. The blank is in units of $\times 10^2 \, \text{mA}$ , so convert $1500 \, \text{mA}$ to $15 \times 10^2 \, \text{mA}$ .

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