The total electrostatic energy stored in both the capacitors (in μJ ) is

Question

The total electrostatic energy stored in both the capacitors (in μJ ) is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-hqfhsauanrad5ktd.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-hqfhsauanradpi8j.jpg" /> According to circuit diagram, capacitors are connected in series combination. So, equivalent capacitance (Ce q)=(3 × 6/3 + 6) =(18/9)=2 μ F We know that, Total electrostatic energy (U)=(1/2)Ce qV2 =(1/2)× 2× (- 3)2 =9 μ J

Q. The total electrostatic energy stored in both the capacitors (in $μ J$ ) is

According to circuit diagram, capacitors are connected in series combination.
So, equivalent capacitance $(C_{e q}) = \frac{3 \times 6}{3 + 6}$
$= \frac{18}{9} = 2 μ F$
We know that,
Total electrostatic energy $(U) = \frac{1}{2} C_{e q} V^{2}$
$= \frac{1}{2} \times 2 \times (- 3)^{2}$
$= 9 μ J$

Q. The total electrostatic energy stored in both the capacitors (in μJ ) is

According to circuit diagram, capacitors are connected in series combination. So, equivalent capacitance (Ceq​)=3+63×6​ =918​=2μF We know that, Total electrostatic energy (U)=21​Ceq​V2 =21​×2×(−3)2 =9μJ

Q. The total electrostatic energy stored in both the capacitors (in $μ J$ ) is

According to circuit diagram, capacitors are connected in series combination.
So, equivalent capacitance $(C_{e q}) = \frac{3 \times 6}{3 + 6}$
$= \frac{18}{9} = 2 μ F$
We know that,
Total electrostatic energy $(U) = \frac{1}{2} C_{e q} V^{2}$
$= \frac{1}{2} \times 2 \times (- 3)^{2}$
$= 9 μ J$