Question

When a series combination of two uncharged capacitors of capacitors C and 4uf are connected to the battery of 12V , 173uJ energy is drawn from the battery then find the volume of C

Answer 1

Explanation:

The total capacitance C of the series can be found from the formula:

$E = \frac{CU^{2} }{2}$

where E = $300\times 10^{-6}J$ is the energy of the system, U = 15V is the voltage. Thus, obtain:

$C = \frac{2E}{U^{2} }$

the capacitance of the two capacitors in cseries is given as follows:

$\frac{1}{C} = \frac{1}{C_{1} } + \frac{1}{C_{2} }$

C1 = 5 x $10^{-12}$

$C_{2} = \frac{5 * 10^{-12F} }{5 * 10^{-12F}\frac{15V^{2} }{2 * 300 * 10^{-6 } j }-1 }$

= -5 x $10^{-12}J$

The ansver is negative, thus, the given numbers are wrong.

Answer 2

Answer:

The volume of C IS  6 μF

Explanation:

A series combination of two uncharged capacitors of capacitors C and 4uf .

 $C_{1} =?$ and $C_{2} = 4$ μf

 Voltage of battery = 12V

 Energy drawn from the batter = 173 μJ   =173×$10^{-6}$

   Capacitors are connected in series $\frac{1}{C} =\frac{1}{C_{1} } +\frac{1}{C_{2} }$------(1)

                                                                $\frac{1}{C} =\frac{C_{1}+C_{2} }{C_{1} C_{2} }$

   ENERGY(E) = $\frac{1}{2} *C*V^{2}$

                       $C=\frac{2*173*10^{-6}J }{12V^{2} }$

                        $C=2.4*10^{-6} F$

 From equation(1) $\frac{1}{C_{1} } =\frac{1}{C}-\frac{1}{C_{2} }$

                            $C_{1} =\frac{C_{2}C }{C_{2}-C }$

                            $C_{1} =\frac{2.4 *4}{4-2.4 }$

                           $C_{1} =6$ μF

   The volume of C IS  6 μF

The project code is #SPJ3