# Answer in Electric Circuits for parrish #159600

Let’s first find the charge on each capacitors which they acquire when charged as a parallel combination across a 250 V battery:

“Q_1=C_1V=6.0cdot10^{-6} Fcdot250 V=1.5cdot10^{-3} C,”“Q_2=C_2V=2.0cdot10^{-6} Fcdot250 V=0.5cdot10^{-3} C.”

Then, the two capacitors are connected positive plate to negative plate and negative plate to positive plate. Let’s find the net charge on the circuit:

“Q_{net}=Q_1-Q_2=1.5cdot10^{-3} C-0.5cdot10^{-3} C=1.0cdot10^{-3} C.”

Let’s find the equivalent capacitance of the parallel combination of two capacitors:

“C_{eq}=C_1+C_2=6.0cdot10^{-6} F+2.0cdot10^{-6} F=8.0cdot10^{-6} F.”

Then, we can find the new voltage across each capacitor (the voltage across each element is the same in parallel circuit):

“V_{new}=dfrac{Q_{net}}{C_{eq}}=dfrac{1.0cdot10^{-3} C}{8.0cdot10^{-6} F}=125 V.”

Finally, we can find the new charge on each capacitor:

“Q_{1,new}=C_1V_{new}=6.0cdot10^{-6} Fcdot125 V=750 mu C,”“Q_{2,new}=C_2V_{new}=2.0cdot10^{-6} Fcdot125 V=250 mu C.”

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