Observability of a wheatstone bridge with capacitor replacing the galvanometer?
Answer:
Think about what the galanometer is measuring -- voltage differential. And it takes a voltage differential to charge a capacitor.
As you tweaked the variable resistance to create a balanced circuit, you'd eventually see a zero charge on the capacitor when you reached circuit balance. Both sides of the capacitor would be seeing the same voltage on each side of the bridge, and there would be no differential (zero current) to charge the capacitor.
I'm not sure what the usefulness would be, but that's what you'd get.
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