In the given figure, there is a circuit of potentiometer of length AB = 10m. The resistance per unit length is 0.1Ω per cm. Across AB, a battery of emf E and internal resistance 'r' is connected. The maximum value of emf measured by this potentiometer is : (The diagram shows a primary circuit with a 6V battery and a 20Ω resistor connected across the potentiometer wire AB. The secondary circuit shows an emf source being measured.)
Maximum value of emf is measured when pointer is at B with I? = 0. So, I_AB = 6 / (20 + 0.1 × 1000) = 6/120 = 1/20 A V_AB = E = I_AB × R_AB = (1/20) × 100 = 5V
<p>Maximum value of emf is measured when pointer is at B with I? = 0.<br>So, I_AB = 6 / (20 + 0.1 × 1000) = 6/120 = 1/20 A<br>V_AB = E = I_AB × R_AB<br>= (1/20) × 100<br>= 5V</p>
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