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0 votes
16.3k views
in Physics by (70.0k points)

Four capacitors are connected as shown in the adjoining fig. The potential difference between A and B is 1500 volt. The energy stored in 2 μF capacitor will be :

(a) 50.6 J 

(b) 0.81 J 

(c) 0.506 J 

(d) none

1 Answer

+1 vote
by (71.8k points)
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Best answer

Correct option (b) 0.81 J

Explanation:

Since, 2 μF and 3 μF capacitors are in parallel. 

So, their effective capacitance is 

C′ = 2 + 3 = 5 μF 

Now 12 μF, 20 μF and C′ are in series. 

Therefore, their effective capacitance is

The charge at C′, 20 μF and 12 μF capacitors is equal 

Hence, potential difference across 2 μF capacitor

So, energy stored in 2 μF capacitor will be

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