Today’s agenda: Measuring Instruments: ammeter, voltmeter, ohmmeter. You must be able to calculate currents and voltages in circuits that contain “real” measuring instruments. RC Circuits. You must be able to calculate currents and voltages in circuits containing both a resistor and a capacitor. You must be able to calculate the time constant of an RC circuit, or use the time constant in other calculations.

RC Circuits RC circuits contain both a resistor R and a capacitor C (duh). Until now we have assumed that charge is instantly placed on a capacitor by an emf. The approximation resulting from this assumption is reasonable, provided the resistance between the emf and the capacitor being charged/discharged is small. If the resistance between the emf and the capacitor is finite, then the capacitor does not change instantaneously. Q t Q t

Switch open, no current flows. Charging a Capacitor  R switch C t<0 Close switch, current flows. t>0 I Apply Kirchoff’s loop rule* (green loop) at the instant charge on C is q. *Convention for capacitors is “like” batteries: negative if going across from + to -. This equation is deceptively complex because I depends on q and both depend on time q +q

When t=0, q=0 and I 0 =  /R. Limiting Cases  R switch C When t is “large,” the capacitor is fully charged, the current “shuts off,” and Q=C . I  = IR is true only at time t=0! V R = IR is always true, but V R is the potential difference across the resistor, which you may not know. Using V = IR to find the voltage across the capacitor is likely to lead to mistakes unless you are very careful

Math:

More math:

Still more math:  = RC is the “time constant” of the circuit; it tells us “how fast” the capacitor charges and discharges. Why not just solve this for q and I?

Charging a capacitor; summary: Sample plots with  =10 V, R=200 , and C=1000  F. RC=0.2 s recall that this is I 0, also called I max

In a time t=RC, the capacitor charges to Q(1-e -1 ) or 63% of its capacity… RC=0.2 s …and the current drops to I max (e -1 ) or 37% of its maximum.  =RC is called the time constant of the RC circuit

Capacitor charged, switch open, no current flows. Discharging a Capacitor R switch C t<0 Close switch, current flows. t>0 I Apply Kirchoff’s loop rule* (green loop) at the instant charge on C is q. *Convention for capacitors is “like” batteries: positive if going across from - to +. +Q -Q-q +q

Math: negative because charge decreases

More math: same equation as for charging

Disharging a capacitor; summary: Sample plots with  =10 V, R=200 , and C=1000  F. RC=0.2 s

In a time t=RC, the capacitor discharges to Qe -1 or 37% of its capacity… RC=0.2 s …and the current drops to I max (e -1 ) or 37% of its maximum.

Notes This is for charging a capacitor.  /R = I 0 = I max is the initial current, and depends on the charging emf and the resistor. This is for discharging a capacitor. I 0 = Q/RC, and depends on how much charge Q the capacitor started with. I 0 for charging is equal to I 0 for discharging only if the discharging capacitor was fully charged.

Notes In a series RC circuit, the same current I flows through both the capacitor and the resistor. Sometimes this fact comes in handy. In a series RC circuit, where a source of emf is present (so this is for capacitor charging problems)... Any technique that begins with a starting equation and is worked correctly is acceptable, but I don’t recommend trying to memorize a bunch of special cases. Starting with I(t) = dq(t)/dt always works. V c and I must be at the same instant in time for this to work.

Notes In a discharging capacitor problem... So sometimes you can “get away” with using V = IR, where V is the potential difference across the capacitor (if the circuit has only a resistor and a capacitor). Rather than hoping you get lucky and “get away” with using V = IR, I recommend you understand the physics of the circuit!

Homework Hints This is always true for a capacitor. Q final = C , where  is the potential difference of the charging emf. Q 0 is the charge on the capacitor at the start of discharge. Q 0 = C  only if you let the capacitor charge for a “long time.” Ohm’s law applies to resistors, not capacitors. Can give you the current only if you know V across the resistor. Safer to take dq/dt.