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In the arrangement shown, a potential difference is applied between points A and B. No capacitor can withstand a potential difference of more than 100 V <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_DPP_PHY_XII_E01_704_Q01.png" width="80%">

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Answer Text

The magnitude of the maximum potential difference that can exist between points A and B is 300 VThe maximum potential difference that can exist across `C_(2)` si 50 VThe maximum charge that can be stored by `C_(3)` is 2 mCThe maximum energy that can be stored in the three capacitor arrangement is 190 mJ

Answer :
B::C::D
Solution :
All capacitors are in series, therefore charge on each capacitor will be same, when connected across a potential source. The capacitor with minimum capacitance will have maximum potential difference across it <br> (evident from `V = (Q)/(C) , Q` being same for all) <br> So 100 V potential difference can exist across `C_(1)`, then charge on it would be `100 V xx 20 mu F = 2000 muC` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_DPP_PHY_XII_E01_704_S01.png" width="80%"> <br> `{:(2000 muC,2000muC,2000 muC),(V_(1)=100V,V_(2)=50 V,V_(3) =40 V),(E_(1)=(Q^(2))/(C_(1)),E_(2)=(Q^(2))/(C_(2)),E_(3)=(Q^(2))/(C_(3))):}` <br> `E = E_(1)+E_(2)+E_(3) = 190 mJ`

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