Uh ... you may want to rethink that.Tofuwu6 said:
Wouldn't the current be 0 and ΔV = IR so ΔV is also 0?phinds said:
Let's say you have a simple circuit with a voltage source of 10V that runs through a resistor of 10 ohms and then through a broken part of wire, creating an open in the circuit. What is the voltage across the open?Tofuwu6 said:
Consider the entire 3 ohm, 4 mu-f leg to be an open circuit (which it IS, since as you point out, the cap is a blockage to current at t = infinity). What is the voltage across that open circuit?Tofuwu6 said:
Ohhhhhhhhhhhhh, so the current of the circuit after the 3 ohm, 4 mu-f leg is blocked off is 3 because I = ΔV/R = 42/(8+6) = 3 then you find the voltage across the 6 ohm resistor which is 6 * 3 = 18V and since this leg is parallel to the leg with the capacitor it would take the same amount of voltage to hold Kirchoff's voltage loop rule true. Then CΔV = 18(4E-6) = 7.2E-5C. Thanks so much!phinds said:
The equation you are referring to doesn't apply to the given circuit. It only applies to the basic RC circuit—a battery, a resistor R, and a capacitor C connected in series.Tofuwu6 said:
That's the voltage across the resistor, not the capacitor. Because there's no potential drop across the 3-ohm resistor, the voltage across the capacitor is the same as the voltage across the 6-ohm resistor.Tofuwu6 said:
My apologies, this was my first time posting anything and this is all the question gave meOrodruin said:
Thanks for clearing up that one bit of information.Tofuwu6 said:
Orodruin said:
The final charge \(Q_f\) on a capacitor in an RC circuit is given by the formula \(Q_f = C \cdot V\), where \(C\) is the capacitance of the capacitor and \(V\) is the voltage across the capacitor after a long period of time (steady-state).
Resistors affect the charging time of a capacitor by determining the time constant \(\tau\) of the circuit, which is given by \(\tau = R \cdot C\), where \(R\) is the resistance and \(C\) is the capacitance. The time constant \(\tau\) represents the time it takes for the capacitor to charge to about 63% of its final value. The larger the resistance, the longer it takes for the capacitor to charge.
The time constant \(\tau\) in an RC circuit is the product of the resistance \(R\) and the capacitance \(C\). It is a measure of the time it takes for the voltage across the capacitor to reach approximately 63% of its final value during charging, or to fall to approximately 37% of its initial value during discharging.
The voltage \(V(t)\) across a capacitor at any time \(t\) during charging in an RC circuit is given by the equation \(V(t) = V_f \left(1 - e^{-\frac{t}{\tau}}\right)\), where \(V_f\) is the final voltage across the capacitor, \(t\) is the time, and \(\tau\) is the time constant of the circuit.
When the circuit reaches steady-state, the capacitor is fully charged and the voltage across it is equal to the supply voltage \(V\). The charge \(Q\) on the capacitor is then given by \(Q = C \cdot V\), where \(C\) is the capacitance of the capacitor. At this point, the current in the circuit drops to zero as there is no further change in the voltage across the capacitor.