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Question
A potentiometer wire of length 1.0 m has a resistance of 15 Ω. It is connected to a 5 V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 60 cm.
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Solution
From the figure:
Total resistance of the circuit , R = (RAB + 5) Ω = 20 Ω
Current in the circuit,
\[i = \frac{V}{R} = \frac{5}{20} A\]
∴ Voltage across AB, VAB = i.RAB = 3.75 V
The emf of the cell connected as above is given by:
\[e = \frac{l}{L} V_0\]
Here, l = 60 cm (balance point)
L = 1 m = 100 cm (total length of the wire)
\[\therefore e = \frac{60}{100}\left( 3 . 75 \right) = 2 . 25 V\]
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