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A capacitor of 20mu F and charged to 200V is connected parallel with parallel with another capacitor of 10muF and charged to 400V. Find the common potential. |
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Answer» SOLUTION :Data supplied, `C_(1)=20 muF, =20 xx 10^(-6)F, C_(2)=10muF=10 xx 10^(-6)F, V_(t)=200V, " "V_(2)=400V` `Q_(1)=C_(1)V_(1), Q_(2)=C_(2)V_(2) and Q=CV` `C=C_(1)+C_(2)` `V=Q/C =(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))` Common potential, `V=((20 xx 200+10 xx 400) xx 10^(-6))/((20+10) xx 10^(-6)) =(4000 +4000)/(30)=(8000)/(30)=266.7"volts"` |
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