CBSE Class 12-science Answered
Please explain this question as early as possible
Asked by modi72879 | 22 Feb, 2018, 06:20: PM
Expert Answer
Left side figure illustrates when the key is open.
Parallel combination of two capacitors of value C each gives effective capacitance 2C. When this 2C capcitance is in series with another capacitor of value C, then the combined effective capacitance is (2/3)C. Hence the charge developed across the effective capacitance q1 which is equal to (2/3)C×V, where V is the voltage of battery. Figure shows the charge value in the capacitor plates.
Right side figure illustrates the condition when key is closed.
closing the key makes the parallel capacitors shorted and the charges will be neutralised. Now after closing the key the effective capacitance becomes C and the charge difference is q2-q1 = (1/3)CV. Hence the charge flow through the battery after closing the key is (1/3)CV columbs
Answered by Thiyagarajan K | 22 Feb, 2018, 09:39: PM
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