Tunable capacitor

If current be constant in the circuit, it implies that the voltage across the $3 \Omega$ resistor must be constant. Hence $V$ of capacitor must be constant and should not drop.
Now as we know that for a capacitor,
$Q=C \times V$ , Therefore
Differentiating w.r.t time both sides, we get
$\frac{\mathrm d}{\mathrm d t} \left( Q \right) = V \times \frac{\mathrm d}{\mathrm d t} \left( C \right)$ , as V is constant.
Now clearly $\frac{\mathrm d}{\mathrm d t} \left( Q \right) = I$ (current in the circuit).
Hence, $\frac{I}{V} = \frac{\mathrm d}{\mathrm d t} \left( C \right) = \frac{1}{R} = \frac{1}{3}$ . Now as we know that in changing the capacitance of the capacitor by varying the distance between plates of capacitor will require some work that would contribute to capacitor's Energy completely.
Therefore, Rate of Change of Energy of Capacitance = External Power Needed.
$\Rightarrow \frac{\mathrm d}{\mathrm d t} \left( \frac{1}{2} CV^2 \right) = P_{required}$
As $V$ is constant hence
$P_{required} = \frac{V^2}{2} \frac{\mathrm d}{\mathrm d t} \left( C \right) = \frac{12 \times 12}{2 \times 3} = 24$

If the current is constant, then

• The voltage given by the capacitator is constant.

• The power lost by the resistor is constant.

The power lost by the resistor per second is given by:

$P = \frac {V^2}{R}$ , which equals 48 from the givens.

The capacitor also gives off constant power, $\frac {U}{1 second}$ , where U is the energy given off by the capacitor.

If one subtracts this from the power lost, you will find the external power needed to keep the voltage constant, $P = 48 - \frac {U}{1 second}$ .

One of the equations for U is,

$U = \frac {1}{2}CV^2$

When we substitute the definition of C in,

$U = \frac {1}{2} (\frac {Q}{V}) V^2 = \frac {1}{2} QV$

When we write it in terms of the current, I,

$\frac {U}{1 second} = \frac {1}{2} I V$ .

Using the definition of Voltage, $V = IR$ . Rearranging,

$I = \frac {V}{R} = \frac {12}{3} = 4$ .

Plugging this back in, $\frac {U}{1 second} = \frac {1}{2} I V = \frac {1}{2} (4) (12) = (2) (12) = 24$ .

Therefore, $P = 48 - 24 = 24 W$

Let the charge on the capacitor at a given instant be q , capacitance of the capacitor be C and the current in the circuit ( which is constant) be \frac {dq}{dt} Using Kirchoff"s Voltage rule in the circuit, we have, \frac {q}{C} = \frac{dq}{dt} \times R , differenciating with respect to time,

\frac{dI}{dt} \times R\ times C^2 = C \frac{dq}{dt} - q \frac{dC}{dt} = 0
which gives, [ where * I = \frac{dq}{dt} * ] C \frac{dq}{dt} = q \frac{dC}{dt}

substituting the value of frac{dg}{dt},

\frac{dC}{dt} = \frac{1}{ R }

Power = F \cdot V= - \frac {dU}{dx} \cdot \frac{dx}{dt}

                           = - **\frac{dU}{dt}**

                          = -\frac{1}{2} \times**C** \times**V^2**
                            **\frac{1}{2} C \times 2 \times V \times \frac {dv}{dt} +       V^2 \times \frac{dC}{dt}**
                         = **\frac {V^2}{2} \times \frac{dC}{dt}**
                          = \frac{V^2}{2R} = **24** watts

Since the resistor and the capacitor are connected in parallel we have from Kirchhoff Law that $U_{C}= \frac{Q(t)}{C(t)}=I R=\textrm{constant}=U_{0}=12~\mbox{V}.$ On the other hand, the constant current flowing in the circuit is just $I=-\frac{dQ(t)}{dt}.$ Combining these two equations we obtain $\frac{dC(t)}{dt}= -\frac{1}{R}.$ Now, the energy stored in the capacitor is given by the formula $E_{p}= \frac{1}{2} C(t) U_{0}^2$ and since $\frac{dC(t)}{dt}<0$ , it may seem that the power needed to keep the current constant is negative. However, we must take into account the power dissipated in the resistor which is given by the formula $I^{2}R$ . Thus, the energy balance equation of the circuit assumes the form $P=I^{2}R+\frac{dE_{p}}{dt}=\frac{U_{0}^2}{R}+ \frac{1}{2}\frac{dC(t)}{dt}U_{0}^2=\frac{U_{0}^{2}}{2R}=24 ~\mbox{W}.$