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Announcements  Special Homework #6 is due tomorrow. You can download it here, if necessary: Today is our last day on circuits. By the end of today, you should have mastered the elementary aspects of dc circuits, including debugging and fixing broken ones.

Contact me by the end of next Wednesdays lecture if you have special circumstances different than for exam 1.  Exam 2 is two weeks from yesterday. Announcements  A note on homework…

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.

You know how to calculate the current in this circuit: Measuring Instruments: Ammeter V R If you don’t know V or R, you can measure I with an ammeter. To minimize error the ammeter resistance r should very small. Any ammeter has a resistance r.Any ammeter has a resistance r. The current you measure is r

Example: an ammeter of resistance 10 m  is used to measure the current through a 10  resistor in series with a 3 V battery that has an internal resistance of 0.5 . What is the percent error caused by the nonzero resistance of the ammeter? V=3 V R=10  r=0.5  Actual current: You might see the symbol  used instead of V.

V=3 V R=10  r=0.5  Current with ammeter: RARA Not bad in a Physics 24 lab!

A Galvanometer When a current is passed through a coil connected to a needle, the coil experiences a torque and deflects. See the link below for more details.

An ammeter (and a voltmeter—coming soon) is based on a galvanometer. OK, everything is electronic these days, but the principles here still apply. We’ll learn about galvanometers later. For now, all you need to know is that the deflection of the galvanometer needle is proportional to the current in the coil (red). A typical galvanometer has a resistance of a few tens of ohms. Hold it right there. Didn’t you say an ammeter must have a very small resistance. Is there a physics mistake in there somewhere?

A galvanometer-based ammeter uses a galvanometer and a shunt, connected in parallel: A I G RGRG R SHUNT IGIG I SHUNT I  Everything inside the blue box is the ammeter. The resistance of the ammeter is

Homework hint: “the galvanometer reads 1A full scale” means a current of I G =1A produces a full-scale deflection of the galvanometer needle. The needle deflection is proportional to the current I G. If you want the ammeter shown to read 5A full scale, then the selected R SHUNT must result in I G =1A when I=5A. In that case, what are I SHUNT and V AB (=V SHUNT )? G RGRG R SHUNT IGIG I SHUNT I Homework hint: “the galvanometer reads 1A full scale” means a current of I G =1A produces a full-scale deflection of the galvanometer needle. The needle deflection is proportional to the current I G. If you want the ammeter shown to read 5A full scale, then the selected R SHUNT must result in I G =1A when I=5A. In that case, what are I SHUNT and V AB (=V SHUNT )? AB Homework hint: “the galvanometer reads 1A full scale” means a current of I G =1A produces a full-scale deflection of the galvanometer needle. The needle deflection is proportional to the current I G. If you want the ammeter shown to read 5A full scale, then the selected R SHUNT must result in I G =1A when I=5A. In that case, what are I SHUNT and V AB (=V SHUNT )?

A galvanometer-based ammeter uses a galvanometer and a shunt, connected in parallel: G RGRG RSRS IGIG ISIS I Example: what shunt resistance is required for an ammeter to have a resistance of 10 m , if the galvanometer resistance is 60  ? The shunt resistance is chosen so that I G does not exceed the maximum current for the galvanometer and so that the effective resistance of the ammeter is very small. (actually  )

To achieve such a small resistance, the shunt is probably a large-diameter wire or solid piece of metal. Web links: ammeter design, ammeter impact on circuit, clamp-on ammeter (based on principles we will soon be studying).ammeter designammeter impact on circuit clamp-on ammeter

Measuring Instruments: Voltmeter You can measure a voltage by placing a galvanometer in parallel with the circuit component across which you wish to measure the potential difference. V=3 V R=10  r=0.5  G RGRG a b V ab =?

Example: an galvanometer of resistance 60  is used to measure the voltage drop across a 10 k  resistor in series with a 6 V battery and a 5 k  resistor (neglect the internal resistance of the battery). What is the percent error caused by the nonzero resistance of the galvanometer? First calculate the actual voltage drop. V=6 V R 1 =10 k  R 2 =5 k  a b

The measurement is made with the galvanometer. V=6 V R 1 =10 k  R 2 =5 k  G R G =60  a b 60  and 10 k  resistors in parallel are equivalent to an 59.6  resistor. The total equivalent resistance is , so 1.19x10 -3 A of current flows from the battery. I=1.19 mA The voltage drop from a to b is then measured to be 6-(1.19x10 -3 )(5000)=0.07 V. The percent error is. Your opinions? Would you pay for this voltmeter?

To reduce the percent error, the device being used as a voltmeter must have a very large resistance, so a voltmeter can be made from galvanometer in series with a large resistance. V G RGRG R Ser  Everything inside the blue box is the voltmeter. a b V ab a b Homework hints: “the galvanometer reads 1A full scale” would mean a current of I G =1A would produce a full-scale deflection of the galvanometer needle. If you want the voltmeter shown to read 10V full scale, then the selected R Ser must result in I G =1A when V ab =10V. Homework hints: “the galvanometer reads 1A full scale” would mean a current of I G =1A would produce a full-scale deflection of the galvanometer needle. If you want the voltmeter shown to read 10V full scale, then the selected R Ser must result in I G =1A when V ab =10V.

Example: a voltmeter of resistance 100 k  is used to measure the voltage drop across a 10 k  resistor in series with a 6 V battery and a 5 k  resistor (neglect the internal resistance of the battery). What is the percent error caused by the nonzero resistance of the voltmeter? We already calculated the actual voltage drop (3 slides back). V=6 V R 1 =10 k  R 2 =5 k  a b

The measurement is now made with the voltmeter. V=6 V R 1 =10 k  R 2 =5 k  V R V =100 k  a b 100 k  and 10 k  resistors in parallel are equivalent to an 9090  resistor. The total equivalent resistance is , so 4.26x10 -4 A of current flows from the battery. I=.426 mA The voltage drop from a to b is then measured to be 6-(4.26x10 -4 )(5000)=3.9 V. The percent error is. Not great, but much better. Larger R ser is needed for high accuracy.

An ohmmeter measures resistance. An ohmmeter is made from a galvanometer, a series resistance, and a battery. G RGRG R Ser R=? The ohmmeter is connected in parallel with the unknown resistance with external power off. The ohmmeter battery causes current to flow, and Ohm’s law is used to determine the unknown resistance.  Measuring Instruments: Ohmmeter Everything inside the blue box is the ohmmeter.

To measure a really small resistance, an ohmmeter won’t work. Solution: four-point probe. V A reference: Measure current and voltage separately, apply Ohm’s law.

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.

Example: For the circuit shown C = 8 μF and ΔV = 30 V. Initially the capacitor is uncharged. The switch S is then closed and the capacitor begins to charge. Determine the charge on the capacitor at time t = 0.693RC, after the switch is closed. (From a prior test.) Also determine the current through the capacitor and voltage across the capacitor terminals at that time. To be worked at the blackboard in lecture.

Demo Charging and discharging a capacitor. Instead of doing a physical demo, if I have time I will do a virtual demo using the applet linked on the next slide. The applet illustrates the same principles as the physical demo.

make your own capacitor circuits For a “pre-built” RC circuit that lets you both charge and discharge (through separate switches), download this file, put it in your “my documents” folder, run the circuit construction applet (link above), maximize it, then select “load” in the upper right. Click on the “capacitor_circuit” file and give the program permission to run it. You can put voltmeters and ammeters in your circuit. You can change values or R, C, and V. Also, click on the “current chart” button for a plot of current (you can have more than one in your applet) or the “voltage chart” button for a plot of voltage. this file

more applets