An ideal battery of emf 2 V and a series resistance R are connected in the primary circuit of a potentiometer of length 1 m and resistance 5 Ω . The value of R , to give a potential difference of 5 mV across 10 cm of potentiometer wire is

Question

An ideal battery of emf 2 V and a series resistance R are connected in the primary circuit of a potentiometer of length 1 m and resistance 5 Ω . The value of R , to give a potential difference of 5 mV across 10 cm of potentiometer wire is, 1 8 0 Ω, 1 9 0 Ω, 1 9 5 Ω, 2 0 0 Ω

MARKS App · Accepted Answer

I = E R + r = 2 R + 5 A Therefore, the potential difference across the potentiometer wire of lengths L = 100 cm is V = I r = 2 R + 5 × 5 = 10 R + 5 V Given, V = 5 mV = 5 × 10 - 3 V for 10 cm of wire. Hence, we have 5 × 10 - 3 = 1 R + 5 ⇒ R = 195 Ω

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