Question

in the electric circuit power dissipated in the resistance R and 2R at an instance after the switch is closed are 9 watt and 2 watt respectively . what is the rate of increase in the energy stored in the capacitor at this instant?​

Answer 1

Answer:

IR=2R3R×I=2I3&H=I2R

HRH3R=(IRI3R)2×R3R=(23)2×13=427

Explanation:

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Answer 2

Answer:

The rate of increased heat energy stored in the capacitor is $\frac{5}{2} CV^2$.

Explanation:

An ideal capacitor does not dissipate energy supplied to it. It stores the heat energy in the form of an electric field between the conducting surfaces.

The heat energy (U) of the capacitor is given as,

$U=\frac{1}{2} CV^2$             (1)

Where,

C=capacity of the capacitor

V=potential difference across the plates of the capacitor

Here in this case let the capacity of the capacitor be "C".

We know that power is given as

$P=\frac{V^2}{R}$           (2)

As per the question we have

P₁=9 Watt

P₂=2Watt

R₁=1

R₂=2R

Now,

By putting the values of P₁ and R₁ in equation (2) we get;

$V_1^2=9$

$V_1=3V$                   (3)

By placing the value of V₁ in equation (1) we get;

$U_1=\frac{1}{2} C \times 3V^2$

$U_1=\frac{9}{2}CV^2$           (4)

By putting the values of P₂ and R₂ in equation (2) we get;

$V_2^2=4$

$V_2=2V$

By placing the value of V₂ in equation (1) we get;

$U_2=\frac{1}{2} C\times2V^2$

$U_2=\frac{4}{2} CV^2$            (5)

The rate of increase in the energy stored in the capacitor is,

$\Delta U=U_1-U_2$        (6)

By substituting equations (4) and (5) in equation (6) we get;

$\Delta U=\frac{9}{2}CV^2-\frac{4}{2} CV^2$

$\Delta U=\frac{5}{2} CV^2$

Hence, the rate of increased heat energy stored in the capacitor is $\frac{5}{2} CV^2$.