# Answer in Electric Circuits for Ulrich #159594

(a) The curcuit diagram of the arrangement described above:

(b) The total charge stored in circuit:

“q=CV_b,”

where *C* – the equivalent capacitance of the circuit:

“C=frac{C_1C_2}{C_1+C_2}+C_3.”

So, “q=V_b(frac{C_1C_2}{C_1+C_2}+C_3)=12(frac{4cdot 10^{-6}cdot 3cdot 10^{-6}}{4cdot 10^{-6}+3cdot 10^{-6}}+6cdot 10^{-6})approx 9.3cdot 10^{-5}space C.”

The charge stored in C3:

“q_3=C_3V_b=6cdot 10^{-6}cdot12=7.2cdot 10^{-5}space C.”

The charges stored in two capacitors C1 and C2, connected in series:

“q_1=q_2=frac{q_{12}}{2},”

“q_{12}=C_{12}V_b,”

where “C_{12}” – the equivalent capacitance of two capacitors, connected in series:

“C_{12}=frac{C_1C_2}{C_1+C_2}.”

So, “q_1=q_2=frac{V_bC_1C_2}{2(C_1+C_2)}=frac{12cdot 4cdot 10^{-6}cdot 3cdot 10^{-6}}{2(4cdot 10^{-6}+3cdot 10^{-6})}approx 1.05cdot 10^{-5}space C.”

Answer: the total charge is “9.3cdot 10^{-5}space C”; the charges stored in each capacitor: “q_1=q_2=1.05cdot 10^{-5}space C”, “q_3=7.2cdot 10^{-5}space C”.