Question
A $10.0-\mu \mathrm{F}$ capacitor is charged to 15.0 $\mathrm{V}$ . It is next connected in series with an uncharged $5.00-\mu \mathrm{F}$ capacitor. The series combination is finally connected across a $50.0-\mathrm{V}$ battery as diagrammed in Figure $\mathrm{P} 26.54$ . Find the new potential differences across the $5.00-\mu \mathrm{F}$ and $10.0-\mu \mathrm{F}$ capacitors.
Step 1
0-\mu \mathrm{F}$ capacitor is given by $Q = CV$, where $C$ is the capacitance and $V$ is the voltage. Substituting the given values, we get $Q = 10.0 \mu \mathrm{F} \times 15.0 \mathrm{V} = 150 \mu \mathrm{C}$. Show more…
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A $10.0-\mu \mathrm{F}$ capacitor is charged to $15.0 \mathrm{V} .$ It is next connected in series with an uncharged $5.00-\mu \mathrm{F}$ capacitor. The series combination is finally connected across a 50.0 -V battery as diagrammed in Figure $\mathrm{P} 20.83 .$ Find the new potential differences across the $5.00-\mu \mathrm{F}$ and $10.0-\mu \mathrm{F}$ capacitors after the switch is thrown closed.
A $10.0-\mu \mathrm{F}$ capacitor is charged to 15.0 $\mathrm{V}$ . It is next connected in series with an uncharged $5.00-\mu \mathrm{F}$ capacitor. The series combination is finally connected across a $50.0-\mathrm{V}$ battery as diagrammed in Figure $\mathrm{P} 26.61$ . Find the new potential differences across the noo-\muF and $10.0-\mu \mathrm{F}$ capacitors after the switch is thrown closed.
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