A tunable capacitor (with variable capacitance) is charged by a U
0=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.
from the internet: W = V^2 / R
12v Battery -> C(t) and 3ÃŽ© R in parallel -> back to battery
so the resistor and the capacitor are in parallel and you want to know the overall resistance so you can figure out the power needed? well this website gives some suggestions...but it's physics over my head. if it looks familiar go for it.
i can't post links so just google resistors-and-capacitors-in-parallel and the first link should be the one i saw
i dun get it :(
To find the power needed to keep the current in the circuit constant, we need to understand the relationship between power, current, and resistance.
The power in a circuit is given by the formula: Power = Current^2 * Resistance
In this case, since the current is constant, we can calculate the power by finding the value of the current and resistance.
Given:
Voltage (V) = 12 V
Resistance (R) = 3 Ω
To find the current (I), we can use Ohm's Law: Voltage = Current * Resistance
Rearranging the formula, we get: Current = Voltage / Resistance
Substituting the given values, we have: Current = 12 V / 3 Ω = 4 A
Now that we have the current (I) and resistance (R), we can calculate the power using the formula mentioned above.
Power = Current^2 * Resistance
= (4 A)^2 * 3 Ω
= 16 A^2 * 3 Ω
= 48 A^2 * Ω
Therefore, the power needed to keep the current in the circuit constant is 48 A^2 Ω or 48 Watts.