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 \( \dots \dots \dots \Omega \).
Correct Answer: 2500
Approach Solution - 1
To determine the resistance \( R \) of the wire, we use Ohm's Law: \( V = IR \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance.
The circuit shows that a 10 kΩ resistor is in parallel with \( R \). The total resistance (\( R_t \)) for resistors in parallel is given by:
\( \frac{1}{R_t} = \frac{1}{R} + \frac{1}{10000} \).
From the graph, at 8 V:
\( I = 4 \) mA \( = 0.004 \) A.
Using Ohm's law for 8 V, \( R_t = \frac{8}{0.004} = 2000 \, \Omega \).
Substituting \( R_t = 2000 \, \Omega \) into the parallel formula:
\( \frac{1}{2000} = \frac{1}{R} + \frac{1}{10000} \)
Solving for \( R \):
\( \frac{1}{R} = \frac{1}{2000} - \frac{1}{10000} = \frac{5}{10000} - \frac{1}{10000} = \frac{4}{10000} \)
\( R = \frac{10000}{4} = 2500 \, \Omega \).
The calculated resistance \( R = 2500 \, \Omega \) fits within the expected range (2500,2500).
Approach Solution -2
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 $R = 2500 \, \Omega$.
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