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Current through the battery in the circuit shown in the figure A. immediately after the switch S is closed is `epsilon//R`B. immediately after the switch S is closed is `epsilon //2R`.C. after long time is `epsilon//2R`.D. after long time is `3epsilon//4R` |
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Answer» Correct Answer - A::C a., c. Immediately after closing the switch, the capacitor will behave like conducting wires, so all the resistances between B and G will be short-circuited and current will flow through ABFEG. Hence current `= epsilon//R`. After long time when the capacitors are fully charged, current will go through ABDG. Hence current = `epsilon//2R` . |
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