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Obtain the equivalent capacitance of the network in figure below. For a 300V supply determine the charge and voltage across each capacitor. |
Answer» Solution : `C_("eff")=(200 xx 100 xx 10^(-12) xx 10^(-12))/(10^(-12) (300))=(200)/(3)pF` `Q=CV=(200)/(3) xx 10^(-12) xx 300 =2 xx 10^(-8)C` `V_("100pF") (C_(4))=(2 xx 10^(-8))/(100 xx 10^(-12))=200V` `V_("100pF") (C_(1))=100V` `Q=100 xx 100^(-12)=10^(-8)C` `Q_("200pF")=200 xx 10^(-12) xx 50=10^(-8)C` |
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