Question

In the circuit as shown in the figure the maximum energy stored in capacitor is 0.216mJ then emf
of cell is

Answer 1

Energy =charge stored x avg potential difference

Answer 2

Question:

In the circuit as shown in the figure the maximum energy stored in capacitor is 0.216mJ then emf of cell is. The circuit diagram was given

Answer:

The emf of the cell is 30 V

Explanation:

Given,

Energy stored in the cell = 0.216 mJ

                                        = 0.216 × 10⁻³ J

The Charge of the capacitor = 3 μF =  3 × 10 ⁻⁶

Formulae:

Energy,    $E_{cap} = \frac{1}{2} CV^{2}$   .................................(1)

              $V_{cap}$   = i × R     .....................................(2)

$i = \frac{E}{R+ \frac{3R}{2} }$ , substitute the value of i in equation (2) we get

$V_{cap} = \frac{E}{R+ \frac{3R}{2} } R$

       $= \frac{2 E}{R} R^{}$

       $= \frac{2E}{5}$

$V_{cap} = \frac{2E}{5}$   ........................................(3)

Substitute the values of V, C and $E_{cap}$ in equation (1) we get,

$E_{cap}$ = 0.216 × 10⁻³ = 216 × 10⁻⁶

From equation (1)

216 × 10⁻⁶ = $\frac{1}{2}$  ×  3 × 10 ⁻⁶×$(\frac{2E}{5} )^{2}$

36 × 25 = E²

E = 6 × 5

E = 30 V   ....................................(4)

The emf of the cell is 30 V, when the maximum energy stored in the capacitor is 0.216 mJ

Answer = 30 V

       

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