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Question
A potentiometer wire of length 1 m has a resistance of 5 Ω. It is connected to a 8 V battery in series with a resistance of 15 Ω. Determine the emf of the primary cell which gives a balance point at 60 cm.
Solution
From the figure:
Total resistance of the circuit, R = (RAB + 15) Ω = 20 Ω
Current in the circuit ,
\[i = \frac{V}{R} = \frac{8}{20} A\]
∴ Voltage across AB, VAB = i.RAB = 2 V
The emf of the cell connected as above is given by: \[e = \frac{l}{L} V_0\]
Here:
l = 60 cm (balance point)
AB = L = 1 m = 100 cm (total length of the wire)
\[\therefore e = \frac{60}{100}\left( 2 \right) = 1 . 2 V\]
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