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InterviewSolution
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A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 300 `mu`C. When potential across the capacitor is reduced by 100 V, the charge stored in it becomes 100 `mu`C. Calculate the potential V and the unknown capacitance. What will be the charge stored in the capacitor if the voltage applied had increased by 100V? |
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Answer» (i) Intial voltage, `V_(1)=V` volts, Charge stored, `Q_(1)=300 mu C` `Q_(1)=Cv_(1) " " ` …(i) Charged potential, `V_(2)=V-100V.` `Q_(2)=100 mu C` `Q_(2)=CV_(2) " " ` ...(ii) By dividing (ii) from (i), we get `(Q_(1))/(Q_(2))=(CV_(1))/(CV_(2))rArr (Q_(1))/(Q_(2))=(V_(1))/(V_(2))` `(300)/(100)=(V)/(V-100)` `V=150` volts. `C=(Q_(1))/(V_(1))=(300xx10^(-6))/(150)=2xx10^(-6)F=2 mu F` (ii) If the voltage applied had increased by 120V, then `V_(3)=150 +100=250V` Hence, charge stored in the capacitor, `Q_(3)=CV_(3)=2xx10^(-6)xx250 =500 mu C` |
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