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Standard XII

Physics

Turns Ratio

Potentiometer...

Question

Potentiometer wire length is $10 m$ , having a total resistance of $10 Ω$ . If a battery of emf $2$ volts (of negligible internal resistance) and a rheostat are connected to it them the potential gradient is $20 m V / m$ ; find the resistance imparted through the rheostat

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Solution

$\begin{matrix} K = 20 m V / m V = K . L V = 20 \times 10 = 200 I = V / R = \frac{200}{10} = 20 A I = \frac{E}{R + R^{1}} = \frac{2}{10 + R^{1}}, 20 = \frac{2}{10 + R^{2}} R^{1} = \frac{99}{10} Ω \end{matrix}$

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Potentiometer wire length is 10m, having a total resistance of 10Ω. If a battery of emf 2 volts (of negligible internal resistance) and a rheostat are connected to it them the potential gradient is 20mV/m; find the resistance imparted through the rheostat

K=20mV/mV=K.LV=20×10=200I=V/R=20010=20AI=ER+R1=210+R1,20=210+R2R1=9910Ω

Potentiometer wire length is $10 m$ , having a total resistance of $10 Ω$ . If a battery of emf $2$ volts (of negligible internal resistance) and a rheostat are connected to it them the potential gradient is $20 m V / m$ ; find the resistance imparted through the rheostat

$\begin{matrix} K = 20 m V / m V = K . L V = 20 \times 10 = 200 I = V / R = \frac{200}{10} = 20 A I = \frac{E}{R + R^{1}} = \frac{2}{10 + R^{1}}, 20 = \frac{2}{10 + R^{2}} R^{1} = \frac{99}{10} Ω \end{matrix}$