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$C_{2}=15.00 \mu F_{1} C_{3}=7.80 \mu F_{1}$, and $C_{4}=3.90 \mu F$. (a) Find the potential difference across each capactor and the charge stored in each. (b). The awitch is now cosed. What is the new find potentiel difference across tach capeciter and the new charge stored in each? $c_{1} \quad V_{1}=$ $\mathrm{V}$ $\mathrm{c}_{2}$ $Q_{1}=$ pe $v_{2}=$ $7 v$ c) $v_{3}=\square$ $Q_{3}=\square$ $v-\quad v c$ c. $Q_{a}=$ $x$

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KOFZRX The First Answerer