RC Circuits. Honors Physics Note 002 c Alex R. Dzierba Honors Physics P222 - Spring, 2004

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1 R ircuits Honors Physics Note 2 c Alex R. Dziera Honors Physics P222 - Spring, 24 Introduction This note concerns the ehavior of circuits that include cominations of resistors, capacitors and possily a attery. apacitors onsider two parallel conducting plates of area A and separation d as shown in Figure (a). Such an arrangement is called a capacitor. If a attery of voltage is connected across this arrangement of conductors charge +Q will collect on one plate and equal and opposite charge Q on the other plate. If the voltage is the magnitude of charge is Q = () where is the capacitance of the capacitor. The capacitance depends only on the geometry of the plates (area and separation) and on the nature of the material etween the plates. In SI units where we use couloms for charge and meters for distance the formula for the capacitance of a parallel plate capacitance is given y: = ɛ oa d (2) Here we assume that the space etween the plates is filled with vacuum (air is a good enough approximation). The constant ɛ o, called the permittivity of free space, determines the scale of the force etween two charges separated y a fixed distance and has the value N m 2. The unit of capacitance is the farad determined y F. It turns out that the farad is unwieldy for example if two circular plates are separated y a millimeter, the radius of the plates needed for a F capacitor would e 6 km! Typical capacitors are measured in units of microfarads (µf = 6 F) or picofarads (pf = 2 F). The symol of a capacitor is shown in Figure (). In this course we will calculate the capacitance for various geometries (e.g. parallel plates, coaxial cylinders, parallel wires, etc..). One technique is integrate the energy density stored in the electric field over the volume of the capacitor. The energy density is given y ɛ o E 2 /2 where E is the electric field and is proportional to charge Q.

2 d +Q Q A Q (a) () Figure : (a): Schematic of a parallel plate capacitor the separation of the plates is d and the area is A. () Symol for a capacitor. The voltage across the capacitor is, the capacitance is and charge Q and Q =. Series comination of capacitors The left circuit of Figure 2 shows two capacitors and 2 connected in series and the right circuit shows the equivalent circuit with the series comination replaced y its equivalent capacitor eq. The voltage difference etween the terminals a and is. Because of the conservation of charge the charge on is the same as the charge on 2 and the same as the charge on eq and equal fo Q. The voltage on the each of the two capacitors in series add to give the total voltage. But = Q/, 2 = Q/ 2 and = Q/ eq which leads to: or eq = + 2 (3) eq = (4) Note that the rule for finding the equivalent of series capacitors is similar to the rule for finding the equivalent of parallel resistors. Parallel comination of capacitors The left circuit of Figure 3 shows two capacitors and 2 connected in series and the right circuit shows the equivalent circuit with the parallel comination replaced y its equivalent capacitor eq. The voltage difference etween the terminals a and is. 2

3 a 2 = eq a +Q Q +Q Q +Q Q 2 Figure 2: The left circuit shows two capacitors and 2 connected in series and the right circuit shows the equivalent circuit with the series comination replaced y its equivalent capacitor eq. The voltage difference etween the terminals a and is. 2 a +Q Q +Q 2 Q 2 = a eq +Q Q Figure 3: The left circuit shows two capacitors and 2 connected in parallel and the right circuit shows the equivalent circuit with the parallel comination replaced y its equivalent capacitor eq. The voltage difference etween the terminals a and is. In this case the voltage across is the same as that across 2 and the same as that across eq. But the charge Q on and the charge Q 2 on 2 add to yield the charge Q on eq. This leads to: eq = + 2 (5) Note that the rule for finding the equivalent of parallel capacitors is similar to the rule for finding the equivalent of series resistors. Discharging capacitor In the circuit of Figure 4 shows a capacitor is connected across a resistor R using a switch S. Initially the capacitor carries charge Q. At t = the switch is closed and the capacitor discharges through the resistor. The charge on the capacitor, Q(t), now varies with time as does the current, i(t), through the resistor. Positive charge flows from the positive plate of the capacitor to the negative plate until the net charge on each plate vanishes. The direction of the current flow is also consistent with the voltage charge across the resistor eing the same as the voltage change across the capacitor: Q(t) = i(t)r (6) 3

4 S R + Q R i(t) + Q(t) efore t= after t= Figure 4: A capacitor is connected across a resistor R using a switch S. Initially the capacitor carries charge Q. At t = the switch is closed and the capacitor discharges through the resistor. The charge on the capacitor, Q(t), now varies with time as does the current, i(t), through the resistor. Since i = dq/dt the charge is decreasing explaining the minus sign leading to: dq Q = dt R (7) Integration: Q Q dq Q = R t dt (8) yields: ( ) Q ln = t Q R (9) or: Q(t) = Q o e t/r () where Q o is the initial charge on the capacitor. The charge decreases exponentially. The current also decreases exponentially: i(t) = dq dt = Q o R e t/r = o R e t/r = i o e t/r () Note that the product of R times must have units of s. The so-called R time constant sets the time scale for a capacitor to discharge or as we will see to discharge. The time required for the capacitor to lose one-half its charge is t /2 = ln(2)r =.693R. 4

5 Energy stored in a capacitor learly as the capacitor discharges current flows though the resistor until the capacitor is fully discharged and the current goes to zero. But current flowing through resistor leads to a power dissipation, P(t), that varies with time: P(t) = i 2 (t)r (2) Integrating the power from t = to t = yields the total initial energy stored in the capacitor: P(t)dt = Q2 o R 2 2 e 2t/R Rdt = Q2 o 2 (3) So a capacitor with charge Q and voltage has stored energy that can e expressed as: where we have made use of Q =. E = Q 2 2 = 2 2 = Q (4) 2 A capacitor, resistor and attery in series In Figure 5 a capacitor is connected in series with a resistor R across a attery using a switch S. Initially the capacitor carries zero charge Q. At t = the switch is closed and the capacitor charges through the resistor. The charge on the capacitor, Q(t), now varies with time as does the current, i(t), through the resistor. At t = the charge on the capacitor is zero and so is the voltage difference across the capacitor so when the switch is closed the voltage provided y the attery appears across the resistor so the initial current through the resistor is /R. As the capacitor starts to charge up a voltage appears across the capacitor and the net voltage across the resistor is the attery voltage minus the capacitor voltage which means that the current through the resistor decreases with time. Eventually the capacitor is fully charged so that the voltage across the capacitor is the same as the attery voltage or the final charge is. At this point the voltage across the resistor is zero so no more current flows. To find the time-dependence of the charge Q(t) and current i(t) we start with the expression for the voltage drops in making a complete loop of the circuit on the right hand side of Figure 5: Q ir = (5) Recalling that i = dq/dt you can verify that the solution is given y: 5

6 S R at t = switch S is closed. Initially is uncharged Figure 5: A capacitor is connected in series with a resistor R across a attery using a switch S. Initially the capacitor carries zero charge Q. At t = the switch is closed and the capacitor charges through the resistor. The charge on the capacitor, Q(t), now varies with time as does the current, i(t), through the resistor..8 Q(t)/Q Q(t)/Q or i(t)/i i(t)/i t/r Figure 6: Plot of Q(t)/Q f [lack curve - see eq. 6] and i(t)/i [gray curve - see eq. 7] as a function of t/r for the charging capacitor circuit of Figure 5. Q(t) = ( e t/r ) = Q f ( e t/r ) (6) sustitution of eq. 6 into eq. 5. Indeed we see that Q(t = ) = and Q(t = ) = = Q f. The current is otained y differentiating eq. 6: i(t) = R e t/r = i e t/r (7) from which we see that the initial current is /R = i as expected and eventually the current goes to zero. 6

7 S Q=Q Q= Figure 7: Prolem S 3 Ω all capacitors equal to. µf Figure 8: Prolem 2 Prolems Prolem Fiigure 7 shows two identical capacitors of capacitance. One of the capacitors carries charge Q = Q and Q =. Initially the switch is open and then closed. (a) How much charge exists on each capacitor after the switch has een closed for some time. () What is the total energy efore and after the switch is closed and account for any difference. Prolem 2 Figure 8 shows five capacitors, all identical and with capacitance equal to. µf. The network of capacitors is connected across a series comination of a attery and a Ω resistor when switch S is closed. Initially all the capacitors are uncharged. A t = the switch is closed and sufficient time all capacitors are fully charged. (a) What is the final charge on each capacitor? () While the capacitors were eing charged what was the total energy supplied y the attery? (c) While the capacitors were eing charged what was the total energy dissipated in the resistor? (d) What is the final total stored energy in the capacitors? 7