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You have two capacitors, one is 6. 0 mu F the othe

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Ta ) Solution: Given, The capacitance of the two capacitors are 6.0 μΕ, and 3.0 μF The battery voltage is 9.0 V The equivalence capacitance of the series configuration is, series, C.2 Here, C, is the equivalence capacitance of the series configuration, C1 is the first capacitance C2 is the second capacitance Substitute the values in the above equation. 3x 6x10-12 (3+6)x10 For parallel configuration, C,-G+C (3+6)x10 Therefore, the equivalence capacitance of the series configuration is12 μF, in parallel configuration it is 9 FSubstitute the values 27x10-5 2x3x10-6 121.5 54x10-5 2x6x10 243 ㎕

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