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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. |
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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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