A tunable capacitor (with variable capacitance) is charged by a U0=12 V battery and then is connected in parallel to a R=3 Ω resistor. The capacitance C(t) of the capacitor is controlled so that the current in the circuit remains constant at all times. What is the power (external) in Watts needed to keep the current in the circuit constant? Note that one possible way of controlling the capacitance could be varying the distance between the plates of the capacitor which, in general, requires some work.

To determine the power (external) needed to keep the current in the circuit constant, we first need to understand the relationship between the current, resistance, and capacitance in an RC circuit.

In an RC circuit, the current (I) can be calculated using Ohm's Law:
I = U/R

Where:
- I is the current in Amperes (A)
- U is the voltage across the resistor in Volts (V)
- R is the resistance in Ohms (Ω)

The charge (Q) on a capacitor is given by:
Q = C * U

Where:
- Q is the charge on the capacitor in Coulombs (C)
- C is the capacitance in Farads (F)
- U is the voltage across the capacitor in Volts (V)

The rate of change of charge with respect to time (dq/dt) is equal to the current in the circuit:
dq/dt = I

Since the current is constant, dq/dt is also constant.

Now, let's differentiate both sides of the equation Q = C * U with respect to time:
dQ/dt = C * dU/dt + U * dC/dt

Since U is constant in this scenario (connected to a battery), dU/dt = 0.

Hence, we are left with:
dQ/dt = U * dC/dt

Recall that dq/dt is equal to the current (I), so:
I = U * dC/dt

This equation can be rearranged to express dC/dt in terms of the current:
dC/dt = I / U

To keep the current constant, the capacitance needs to change at a rate that compensates for the changing charge on the capacitor due to the current passing through the resistor.

Now, let's calculate the power (external) needed to keep the current constant. Power is defined as the rate at which work is done:

Power = dW/dt

In this case, the work done is equal to the change in potential energy of the capacitor. The potential energy (Uc) stored in a capacitor is given by:

Uc = (1/2) * C * U^2

Differentiating both sides with respect to time, we get:

dUc/dt = (1/2) * (dC/dt) * U^2 + (1/2) * C * (dU/dt) * 2U

Since dU/dt = 0, we have:

dUc/dt = (1/2) * (dC/dt) * U^2

Now, recall that the power is given by dW/dt. Substituting dUc/dt with the above equation, we have:

Power = (1/2) * (dC/dt) * U^2

From the earlier equation, we found that dC/dt = I / U. Substituting this into the power equation, we get:

Power = (1/2) * (I / U) * U^2

Simplifying, we have:

Power = (1/2) * I * U

Finally, substituting the given values:
I = U / R = 12 V / 3 Ω = 4 A

Power = (1/2) * 4 A * 12 V = 24 W

Therefore, the power (external) needed to keep the current in the circuit constant is 24 Watts.