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RC Circuits. Voltage on the Capacitor Initially there is no voltage across the capacitor, since there is no charge on the plates. The current is large.

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Presentation on theme: "RC Circuits. Voltage on the Capacitor Initially there is no voltage across the capacitor, since there is no charge on the plates. The current is large."— Presentation transcript:

1 RC Circuits

2 Voltage on the Capacitor Initially there is no voltage across the capacitor, since there is no charge on the plates. The current is large and all the voltage drop is across the resistor. As the voltage builds on the capacitor current falls to zero when the capacitor is completely charged.

3 Approaching RC Circuits Create a Kirchoffs loop equation for the circuit. Before we can solve this equation, all variables must be expressed in terms of time and charge. This can be done by relating the voltage across the capacitor to its charge using the equation Q = CV and relating the current through the circuit with the charge on the capacitor using the relationship I = dQ/dt

4 Contd Now we should have a differential equation involving various constants, Q and t. After separating variables, indefinite integration yields an equation relating charge to time with a constant of integration. Use the initial conditions to eliminate the constant of integration. Now we should have charge as a function of time Q(t). If we want to calculate current as a function of time, we can pulg Q(t) into our previous formula relating current and time I = dQ/dt. We should be able to check that our new equation gives the right current values at t = 0 and t = infinity.

5 RC Circuits The current through the circuits discussed so far have been time- independent, as long as the emf of the source is time- independent. The currents in these circuits can be determined by applying Kirchhoff's first and/or second rule. A simple circuit in which the current is time dependent is the RC circuit which consists of a resistor R connected in series with a capacitor C. Applying Kirchhoff's second rule to the current loop I gives ξ – IR – Q/C = 0

6 Current The current in the resistor and the capacitor is the same. I = dq/dt ξ – (dq/dt)R – Q/C = 0 This can be rearranged so that it can be solved in the following manner:

7 Differential Equation dq/(Cξ – Q) = dt/(RC) Now integrate each side dQ/(Cξ - Q) = dt/(RC) -ln (Cξ-Q) +ln(Cξ-Q o ) = ln[(Cξ-Q o )/(Cξ-Q)]=t/(RC) Q o is the charge at time t = 0 QoQo Q QoQo Q

8 Continued… Q = Cξ(1-e -t/RC ) = Q(1-e -t/RC ) Where Q = Cξ is the maximum charge on the capacitor I = (ξ/R)e -t/RC

9 Time Constant, tau RC is the time constant, tau ln (q/Q) = -t/(RC) q = Q o e (-t/RC) I = -dq/dt = I o e (-t/RC) Q o is the initial charge on the capacitor I o = Q/(RC)

10 Problem A 6 microfarad capacitor is charged through a 5 kΩ resistor by a 500 V power supply. How long does it require for the capacitor to acquire 99 percent of its final charge?

11 Answer Tau = RC = 5000Ω ( 6 x F) = sec q =.99Q = Q (1-e -t/RC ) e -t/RC = 0.01 t/RC = -ln (0.01) t/RC = 0.14 sec


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