- The equivalent capacitance between point a and b is 5.87μf.
- The charge at 20μf is 93.92 μC.
- The charge at 6μf is 67.8 μC.
- The charge at C and 3μf is 26.12 μC.
<h3>
Sum of capacitance of C and 3μF</h3>
The sum of the capacitance is calculated as follows;
1/Ct = 1/C + 1/3
1/Ct = 1/10μf + 1/3μf
1/Ct = (3μf + 10μf)/30μf²
1/Ct = 13μf/30μf²
Ct = 30μf²/13μf
Ct = 2.31μf
<h3>Total capacitance in parallel arrangement</h3>
The total capacitance in parallel arrangement is calculated as follows;
Ct = 2.31μf + 6μf = 8.31μf
<h3>Equivalent capacitance between point a and b</h3>
1/Ct = 1/8.31μf + 1/20μf
1/Ct = 0.1703
Ct = 5.87μf
<h3>Charge flowing in each capacitor</h3>
Maximum voltage is delivered in 20μf, q = CV
<u>charge for 20μf:</u>
q = (5.87 x 16)μC
q = 93.92 μC
<h3>Equivalent capacitance for C, 3μf and 6μf</h3>
Ct = 2.31μf + 6μf = 8.31uf
<u>charge for 6μf:</u>
q = (6/8.31) x 93.92μC
q = 67.8μC
<h3>Total charge for C and 3μf</h3>
q = 93.92μC - 67.8μC = 26.12 μC
charge for C = charge 3μf = 26.12 μC
Learn more about capacitor here: brainly.com/question/14883923
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