Question

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Answer 1

I'll explain two methods. The second method may not be applicable to all

problems.

Method 1)

Step 1: Assign charges to each capacitor.

Let charge on 6uF be q1, on 12uF be q2.

Hence, charge on 9uF becomes q1+q2 by charge conservation.

Step 2: Use Kirchhoff's voltage laws.

In the loop of 6 and 12uF, we have:

$\frac{q_1}{6}=\frac{q_2}{12}\\</p><p>\implies 2q_1=q_2$

In the loop of "battery", 6 and 9uF:

$\frac{q_1}{6}+ \frac{q_1+q_2}{9}=90$

Step 3: Solve the two equations to get required charge (q1)=$\boxed{180 \mu C}.$

Method 2)

Step 1: Notice that 6uF and 12uF are in parallel. Hence, net capacitance is 18uF.

Step 2: This 18uF is in series with 9uF.

The net capacitance of the circuit is now 6uF.

Step 3: Let total charge provided be q.

Hence, by voltage law,

$\frac{q}{6}=90\\</p><p>q=540.$

Step 4:

$q_1+q_2=q=540\\</p><p>2q_1=q_2$

Solve the two.

Answer 2

Answer:

THIS IS UR ANSWER

Explanation:

Method 1)

Step 1: Assign charges to each capacitor.

Let charge on 6uF be q1, on 12uF be q2.

Hence, charge on 9uF becomes q1+q2 by charge conservation.

Step 2: Use Kirchhoff's voltage laws.

In the loop of 6 and 12uF, we have:

\begin{gathered} \frac{q_1}{6}=\frac{q_2}{12}\\ < /p > < p > \implies 2q_1=q_2\end{gathered}

6

q

1

=

12

q

2

</p><p>⟹2q

1

=q

2

In the loop of "battery", 6 and 9uF:

\frac{q_1}{6}+ \frac{q_1+q_2}{9}=90

6

q

1

+

9

q

1

+q

2

=90

Step 3: Solve the two equations to get required charge (q1)=\boxed{180 \mu C}.

180μC

.

Method 2)

Step 1: Notice that 6uF and 12uF are in parallel. Hence, net capacitance is 18uF.

Step 2: This 18uF is in series with 9uF.

The net capacitance of the circuit is now 6uF.

Step 3: Let total charge provided be q.

Hence, by voltage law,

\begin{gathered}\frac{q}{6}=90\\ < /p > < p > q=540. \end{gathered}

6

q

=90

</p><p>q=540.

Step 4:

\begin{gathered}q_1+q_2=q=540\\ < /p > < p > 2q_1=q_2\end{gathered}

q

1

+q

2

=q=540

</p><p>2q

1

=q

2

Solve the two.