How to solve the following system of equations for C_0C0 in terms of CC?

V=Q_2/C+Q_3/CV=Q2C+Q3C
V=Q_1/C+(Q-Q_3-Q_4)/CV=Q1C+Q−Q3−Q4C
V=Q_1/C+(Q_1+Q_3+Q_4-Q)/C+(Q_3+Q_4-Q_2)/C+Q_4/CV=Q1C+Q1+Q3+Q4−QC+Q3+Q4−Q2C+Q4C
V=(Q-Q_1-Q_2)/C+(Q_3+Q_4-Q_2)/C+(Q_3-Q_2)/C+Q_3/CV=Q−Q1−Q2C+Q3+Q4−Q2C+Q3−Q2C+Q3C
V=Q_2/C+(Q_2-Q_3)/C+Q_4/CV=Q2C+Q2−Q3C+Q4C
C_0=Q/VC0=QV

PLEASE PROVIDE STEPS.

1 Answer
Cesareo R.
Mar 12, 2018

C_0 = (15 C)/11C0=15C11

Explanation:

Solving the linear system

((0, 0, 1, 1, 0, 0),(1, 1, 0, -1, -1, 0),(-1, 2, -1, 2, 3, 0),(1, -1, -3, 3, 1, 0),(0, 0, 2, -1, 1, 0),(1, 0, 0, 0, 0, -V))((Q),(Q_1),(Q_2),(Q_3),(Q_4),(C_0)) = ((C V),(C V),(C V),(C V),(C V),(0))

we obtain

Q = (15 C V)/11, Q_1 = (5 C V)/11, Q_2 = (6 C V)/11, Q_3 = (5 C V)/11, Q_4 = (4 C V)/11, C_0 = (15 C)/11

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