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Question: A \(4.00\;{{\mu F}}\) capacitor and a \(6.00\;{{\mu F}}\) capacitor are connected in parallel across...

A 4.00  μF4.00\;{{\mu F}} capacitor and a 6.00  μF6.00\;{{\mu F}} capacitor are connected in parallel across a 660  V660\;{\text{V}} supply line. The charged capacitors are disconnected from the line and from each other, and then reconnected to each other with terminals of unlike sign together. Find the final charge on each and the voltage across each capacitor.

Explanation

Solution

The charge on each capacitor will be different when connected in parallel. The charge will be the product of capacitance and the common potential across the capacitors. And when charged capacitors are disconnected and reconnected with unlike terminals the total charge will change.

Complete step by step answer:
Given two capacitors are connected in parallel across a 660  V660\;{\text{V}}. Hence if they are connected parallel the potential difference will be the same for both capacitors. The expression for finding charge on each capacitor is given as,
Q=CVQ = CV
Where, CC is the capacitance and VV is the potential difference across the capacitor.
Therefore the charge on 4.00  μF4.00\;{{\mu F}} is given as,
Q1=CV =4×10−6F×660  V  = 2.64×10−3  C  {Q_1} = CV \\\ = 4 \times {10^{ - 6}}{\text{F}} \times 660\;{\text{V}} \\\ {\text{ = 2}}{\text{.64}} \times {\text{1}}{{\text{0}}^{ - 3}}\;{\text{C}} \\\
And the charge on 6.00  μF6.00\;{{\mu F}}is given as,
Q2=CV =6×10−6F×660  V  = 3.96×10−3  C  {Q_2} = CV \\\ = 6 \times {10^{ - 6}}{\text{F}} \times 660\;{\text{V}} \\\ {\text{ = 3}}{\text{.96}} \times {\text{1}}{{\text{0}}^{ - 3}}\;{\text{C}} \\\
The total capacitance when connected in parallel is given as,
CT=C1+C2 =4  μF+6  μF =10  μF  {C_T} = {C_1} + {C_2} \\\ = 4\;\mu F + 6\;\mu F \\\ = 10\;\mu F \\\
When the capacitors are disconnected and again connected each other with the terminals of unlike sign together, then the total charge is given as,
QT=Q2+(−Q1){Q_T} = {Q_2} + \left( { - {Q_1}} \right)
Substituting the values,
QT=3.96×10−3  C - 2.64×10−3  C =1.32×10−3  C  {Q_T} = {\text{3}}{\text{.96}} \times {\text{1}}{{\text{0}}^{ - 3}}\;{\text{C - 2}}{\text{.64}} \times {\text{1}}{{\text{0}}^{ - 3}}\;{\text{C}} \\\ = 1.32 \times {\text{1}}{{\text{0}}^{ - 3}}\;{\text{C}} \\\
After the reconnection the capacitors are connected in parallel itself. Therefore the common potential difference is given as,
V′=QTCT =1.32×10−3  C10  μF =1.32×10−3  C10×10−6  F =132  V  V' = \dfrac{{{Q_T}}}{{{C_T}}} \\\ = \dfrac{{1.32 \times {\text{1}}{{\text{0}}^{ - 3}}\;{\text{C}}}}{{10\;{{\mu F}}}} \\\ = \dfrac{{1.32 \times {\text{1}}{{\text{0}}^{ - 3}}\;{\text{C}}}}{{10 \times {{10}^{ - 6}}\;{\text{F}}}} \\\ = 132\;{\text{V}} \\\
The final charges on the capacitors are again found after reconnection.
The final charge on 4.00  μF4.00\;{{\mu F}} is given as,
Q′1=CV′ =4×10−6F×132  V  = 5.28×10−4  C  {{Q'}_1} = CV' \\\ = 4 \times {10^{ - 6}}{\text{F}} \times 132\;{\text{V}} \\\ {\text{ = 5}}{\text{.28}} \times {\text{1}}{{\text{0}}^{ - 4}}\;{\text{C}} \\\
And the final charge on 6.00  μF6.00\;{{\mu F}}is given as,
Q′2=CV′ =6×10−6F×132  V  = 7.92×10−4  C  {{Q'}_2} = CV' \\\ = 6 \times {10^{ - 6}}{\text{F}} \times 132\;{\text{V}} \\\ {\text{ = 7}}{\text{.92}} \times {\text{1}}{{\text{0}}^{ - 4}}\;{\text{C}} \\\
Thus the final charges on 4.00  μF4.00\;{{\mu F}} and 6.00  μF6.00\;{{\mu F}} are 5.28×10−4  C{\text{5}}{\text{.28}} \times {\text{1}}{{\text{0}}^{ - 4}}\;{\text{C}}and 7.92×10−4  C{\text{7}}{\text{.92}} \times {\text{1}}{{\text{0}}^{ - 4}}\;{\text{C}} respectively.

Note: We want to note that the total capacitance decreases when the capacitors are connected parallel. And the total charge will be maximum in parallel connection of capacitors when the like terminals of capacitors are connected together.