C_2 and C_3 are uncharged. after the switch is closed, what are the charges on each capacitor? what are the potential difference across each capacitor?

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so before switch is closed, Q=15uF*100V=1500uC found a similar problem online that says to divide the V in half to find individual charges and V after switch is closed. so Q_1=(15uF)100V/2=750uC Q_2+3=1500uC-750uC=750uC Q_2 should equal Q_3 so 750uC/2=375uC V_1 should equal V_2+3: 50V V_2+3=(Q_2+3)/C_2+(Q_2+3)/C_3=(750/20)+(750/30)=62.5V...not equal to 50V

BOth branches are in parallel, so voltage is the same. 100V=Voltage across Cap1=Voltage across Cap2+Voltage across Cap3 For Cap1: Q1=C1V1=(15E-6)*(1E2)=15E-4 C = 150 mC Vcap1=Vcap2+Vcap3 Q1/C1=Q2/C2+Q3/C3 Charge conservation: Q1=Q2+Q3 Then we have two equations with two unknowns easily solvable. I hope it helps.

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