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Three capacitors C_1,C_2,andC_3 of capacitance 1muF, 2muF, and 3muF, respectively, are charged separately as shown in the figure. Now these charged capacitors are connected to a battery of epsilon = 20 V and an uncharged capacitor of C = 2muF as shown in figure. , The potential of point P is |
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Answer» `12.5V`  The circuit can be REDRAWN as Conserving CHARGES on lower plates, we get `-[(X+20)1+(x+2)2+(x+20)3+2x]=-60` or `x=-(15)/(2)V` `|Q_(1muF)|=12.5muC` `|Q_(2MUF)|=25muC` `|Q_(3muF)|=37.5muC` `|Q_(2muF)|=15muC` |
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