Question

CLASS 12th
Ch ELECTROSTATIC POTENTIAL AND CAPACITANCE.

SOLVE WITH PROPER STEPS

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ANSWER->

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Answer 1

Question:

The charge on the $6 \mu F$ capacitor in the circuit is?

Solution:

See the attachment.

The capacitors can be arranged in the way as shown in the attachment.

Now as we can see that $12 \mu F$ and $6 \mu F$ capacitor are connected in parallel.

So we can simply add them i.e $12 \mu F + 6 \mu F= 18 \mu F$

Now, $18 \mu F$ and $9 \mu F$ are in series, so

$Capacitance= \frac{18 \times 9}{18+9} =\frac{162}{27} = \bf 6 \mu F$

Now we have a formula $\boxed{C = \frac{Q}{V}}$

$Q = C \times V \\\\ Q= 6 \mu F \times 90 V = 540 \mu C$

Now since, the charge is same across a parallel set of capacitor, we have $Q= 540 \mu C$

$Since, ~V = \frac{Q}{C} \\\\ V = \frac{540}{18} = 30 V$

Using same formula, we will find charge across $6 \mu F$

$Q= C \times V \\\\ Q = 6\times 30 \\\\ \bf Q=180 \mu C$

Answer 2

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