The total electrostatic energy stored in both the capacitors is :

Question

The total electrostatic energy stored in both the capacitors is : (A)  18 μ J (B) 9 μ J (C) 40.5 μ J (D) 13.5 μ J

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a67bef55edbb2fad90f45d94e3b08f6f-.png" /> According to circuit diagram, capacitors are connected in series combination. So, equivalent capacitance (Ceq)=(3 × 6/3+6) =(18/9)=2 μ F We know that, Total electrostatic energy (U)=(1/2) Ce q V2 =(1/2) × 2 ×(-3)2 =9 μ J

Q. The total electrostatic energy stored in both the capacitors is :

$18 μ J$

$9 μ J$

$40.5 μ J$

$13.5 μ J$

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 is :

18μJ

9μJ

40.5μJ

13.5μJ

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

$18 μ J$

$9 μ J$

$40.5 μ J$

$13.5 μ J$

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 $(U) = \frac{1}{2} C_{e q} V^{2}$
$= \frac{1}{2} \times 2 \times (- 3)^{2}$
$= 9 μ J$