Answer to Question #240946 in Atomic and Nuclear Physics for Asande

Question

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?

Answer 1

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

$$C{eq}=C_1+C_2=5.0\ \mu F+12.0\ \mu F=17.0\ \mu F.$$

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

$$\Delta V_1=\Delta V_2=\Delta V=9.0\ V.$$

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

$$Q_1=C_1\Delta V=5.0\cdot10^{-6}\ F\cdot9.0\ V=45\ \mu C,$$ $$Q_2=C_2\Delta V=12.0\cdot10^{-6}\ F\cdot9.0\ V=108\ \mu C.$$