To determine the resistance (R) of a wire,a c | vedclass

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(B) From the circuit diagram,the voltmeter is connected in parallel with the $10 	ext{ k}\Omega$ resistor and the resistor $R$. However,the voltmeter

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$2500$

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(B) From the circuit diagram,the voltmeter is connected in parallel with the $10 	ext{ k}\Omega$ resistor and the resistor $R$. However,the voltmeter measures the voltage across the parallel combination of $R$ and $10 	ext{ k}\Omega$. Let $R_p$ be the equivalent resistance of $R$ and $10 	ext{ k}\Omega$ in parallel.$R_p = \frac{R 	imes 10^4}{R + 10^4}$From the $V-I$ graph,we can take any point,for example,$V = 4 	ext{ V}$ and $I = 2 	ext{ mA} = 2 	imes 10^{-3} 	ext{ A}$.Using Ohm's law,$V = I R_p$,so $R_p = \frac{V}{I} = \frac{4}{2 	imes 10^{-3}} = 2000 	ext{ } \Omega$.Now,equate the two expressions for $R_p$:$2000 = \frac{R 	imes 10^4}{R + 10^4}$$2000(R + 10000) = 10000R$$2000R + 2 	imes 10^7 = 10000R$$8000R = 2 	imes 10^7$$R = \frac{20000000}{8000} = 2500 	ext{ } \Omega$.

To determine the resistance $(R)$ of a wire,a circuit is designed as shown 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 . . . . . . . $\Omega$.

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