1.

A potentiometer wire of length 1.0 m has a resistance of 15 ohm. It is connected to a 5 V source in series with a resistance of `5 Omega`. Determine the emf of the primary cell which has a balance point at 60 cm.

Answer»

L = 1m = 100cm

r = 15Ω

\(\epsilon\) = 5V

R = 5Ω

\(\epsilon' = \frac{\epsilon r}{R + r} \times \frac rL\)

\(\epsilon' = \frac{5\times 15}{(5 + 15)} \times \frac{60}{100}\)

\(\epsilon' = \frac{4500}{2000}\)

\(\epsilon'= 2.25V\)



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