Question

To determine the resistance \( R \) of a wire, a circuit is designed below. The V-I characteristic curve for this circuit is plotted for the voltmeter and the ammeter readings as shown in the figure. The value of \( R \) is \( \dots \dots \dots \Omega \).

Answer 1

Correct Answer: 2500

The equivalent resistance $R_\text{eq}$ of two resistors in parallel is given by:
 \[ R_\text{eq} = \frac{10^4 R}{10^4 + R}. \] 
Given: \[ E = 4 \, \text{V}, \quad I = 2 \, \text{mA}. \] 
From Ohm's Law: \[ I = \frac{E}{R_\text{eq}}. \] 
Substitute $R_\text{eq}$ into the equation: 
\[ 2 \times 10^{-3} = \frac{4}{\frac{10^4 R}{10^4 + R}}. \] 
Simplify: \[ 2 \times 10^{-3} = \frac{4(10^4 + R)}{10^4 R}. \] 
Multiply through by $10^4 R$: \[ 2 \times 10^4 R = 4(10^4 + R). \] 
Distribute and simplify: \[ 20R = 40000 + 4R. \] 
Rearranging terms: \[ 16R = 40000. \] 
Solve for $R$: \[ R = \frac{40000}{16} = 2500 \, \Omega. \] 
Thus, the resistance $R$ is: \[ \boxed{2500 \, \Omega}. \] 

Explanation: The equivalent resistance for two parallel resistors is determined by the formula $R_\text{eq} = \frac{10^4 R}{10^4 + R}$. By using the provided voltage and current values, Ohm's Law was applied to derive $R$. The algebraic simplifications lead to $R = 2500 \, \Omega$.