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Question
The potential difference across the terminals of a battery of emf 12 V and internal resistance 2 Ω drops to 10 V when it is connected to a silver voltameter. Find the silver deposited at the cathode in half an hour. Atomic weight of silver is 107.9 g mol−1.
Solution
Let i be the current through the circuit.
Emf of battery, E = 12 V
Voltage drop across the voltameter, V = 10 V
Internal resistance of the battery, r = 2 Ω
Applying Kirchoff's Law in the circuit, we get:-
\[E = V + ir\]
\[ \Rightarrow i = \frac{E - V}{r} = \frac{12 - 10}{2} = 1 A\]
Using the formula m = Zit, we get:-
\[m = \frac{107 . 9}{96500} \times 1 \times 0 . 5 \times 3600 = 2 . 01 g\]
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