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Four condensers each of capacity 4 \mu \mathrm{f} are connected as shown in figure \mathrm{ v_p-V_Q=15 volts} The energy stored in the system is -
 

Option: 1

2400 \text { ergs. }


Option: 2

1800 \text { ergs. }


Option: 3

3600 \text { ergs. }


Option: 4

5400 \text { ergs. }


Answers (1)

best_answer

Equaraknt circuit will be-



\mathrm{\begin{aligned} & \frac{1}{C_{P Q}}=\frac{1}{4}+\frac{1}{s}+\frac{1}{4}=\frac{1}{2}+\frac{1}{s}=\frac{4+1}{8} \\ & C_{P a}=\frac{8}{5} \mu \mathrm{F} \end{aligned} }
Now,
store energy 
\mathrm{\begin{aligned} & U=\frac{1}{2} c v^2 \\ & =\frac{1}{2} \times \frac{8}{5} \times 15^3 \times 15 \\ & =12 \times 15=180 \times 10^{-6} \\ & =1800 \text { ergs. } \\ & \end{aligned} }
 

Posted by

Devendra Khairwa

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