Question #240946

Two capacitors, C1 = 5.00 μF and C2 = 12.0 μF, are connected in parallel, and the resulting combination is connected to a 9.00-V battery. (a) What is the equivalent Figure 1 2 capacitance of the combination? What are (b) the potential difference across each capacitor and (c) the charge stored on each capacitor?


1
Expert's answer
2021-09-23T08:30:41-0400

(a) We can find the equivalent capacitance of the combination as follows:


Ceq=C1+C2=5.0 μF+12.0 μF=17.0 μF.C{eq}=C_1+C_2=5.0\ \mu F+12.0\ \mu F=17.0\ \mu F.

(b) Since all capacitors connected in parallel have the same applied potential difference, we can write:


ΔV1=ΔV2=ΔV=9.0 V.\Delta V_1=\Delta V_2=\Delta V=9.0\ V.

(c) We can find the charge stored on each capacitor as follows:


Q1=C1ΔV=5.0106 F9.0 V=45 μC,Q_1=C_1\Delta V=5.0\cdot10^{-6}\ F\cdot9.0\ V=45\ \mu C,Q2=C2ΔV=12.0106 F9.0 V=108 μC.Q_2=C_2\Delta V=12.0\cdot10^{-6}\ F\cdot9.0\ V=108\ \mu C.

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