Current, Voltage measurements;

Potentiometers and bridgesExplain how to extend the range of an ammeter

Explain how to extend the range of an voltmeter

Describe a potentiometer

Explain how an emf can be measured exactly using a driver cell

Outline the Wheatstone Bridge

Current can be measured using a moving coil ammeter or a digital multi-meter, which is put in series with the circuit.

We have a circuit and ammeter. What should we do?Cut the circuit and reconnect it so that the ammeter is

in series:

Current and voltage measurements

Has a known

FSD (Full Scale

Deflection)

Current MeasurementWhat can we do if current is too high: the full scale deflection of the meter is not enough to measure the current?Analogy: to protect us against flood we can dig out a channel to redirect extra water! So we should attach a new wire to allow current flow along this new path (not through our ammeter).

The range of an ammeter may be extended by inserting a shunt resistance in parallel.

Current MeasurementThe range of a meter may be extended by inserting a shunt resistance in parallel.

What should we do if the current is too high? Example: if we remove one bulb from the circuit, then the current will be too high to measure.Indeed the current with two bulbs is already close to the full scale deflection of our meter. This ammeter cannot measure current of a circuit with one bulb only.

?

Current MeasurementThe range of a meter may be extended by inserting a shunt resistance in parallel.

Some current flows through the resistance which is in parallel with the ammeter. Thus the ammeter can measure current which is higher than full scale deflection of the meter. This is because the actual current I is split into two currents, I1 and I2.

I

I1

I2

Current MeasurementHow to calculate what resistance to use?Applying Kirchoff’s current law for node A:I=I1+i. Therefore, I1=I-i

Let the resistance of ammeter be Rm

Applying the Kirchoff’s law to the blue loop below we haveVm-VR=0. Therefore Vm=VR

A

II1

i

I

i

I1=I-i

A

Vm

VR

Current MeasurementFrom the two Kirchoff’s lawsI1=I-i ; Vm=VR

Using Ohm’s Law: Vm= Rmi; VR=R (I-i)

Therefore we have the equationRmi = Vm=VR = R (I-i)

Rmi = R (I-i),

I

i

I1=I-i

A

Vm

VR

iIiR

R m

Current MeasurementCurrent can be measured using a moving coil ammeter or a digital multi-meter, which is put in series with the circuit.The range of a meter may be extended by inserting a shunt resistance in parallel.If i is the full scale deflection (f.s.d.) current of the meter, Rm is the meter resistance and I is the desired new f.s.d. current, then:

iIiR

R m

FP3, questions 5 & 6a(use formula and picture on handout)

Voltage Measurement

A voltmeter is really an ammeter with a series resistance (multiplier resistance).

=

V= I(Rm+R)

R

Rm

Voltage MeasurementVoltmeter is used in parallel and draws only a small current from the circuit.

How to measure voltage across a bulb?

No need to cut the circuit,

just attach the voltmeter in parallel to the bulb. Has a known

FSD (Full Scale

Deflection)

Voltage MeasurementWhat can we do if voltage is too high: the full scale deflection of the voltmeter is not enough to measure voltage?

Let us divide the voltageThe range of the meter may

be varied using the series resistance.

Voltage MeasurementWhat should we do if the voltage is too high? Example: if we want to measure the voltage V2 across two bulbs, then the voltage will be too high to measure.Indeed, the voltage across the two bulbs is about twice as much as the voltage across one bulb, but even with one bulb only the voltage was nearly the full scale deflection.To divide too high a voltage we can add the resistance in series to our voltmeter.

V2

Voltage MeasurementHow to estimate the required resistance to extend the range of the voltmeter from v to V ? Applying the Kirchoff’s law to the green loop:

V2=V =v+VR =v+IR with current

flowing through the resistorSince the voltmeter and the resistor are in series, their resistances should add Rtotal = R+Rm.

Applying Ohm’s Law:I=V/(R+Rm)

Substituting the current I to ,we arrive to the following equation V=v+RV/(R+Rm) Which we now

simplify

V2 =V

R

I

Rm

VR

Ohm’s Law: VR =IR

Voltage MeasurementV=v+RV/(R+Rm)

Multiplying by R+Rm

V (R+Rm) = v (R+Rm)+RV

V R+ V Rm = v R+v Rm +RV

(V -v )Rm = v R

Dividing by v

R=Rm(V-v)/v

V2 =V

R

I

Rm

Voltage MeasurementA voltmeter is really an ammeter with a series resistance (multiplier resistance).The voltmeter is used in parallel and draws only a small current from the circuit.The range of the meter may be varied using the series resistance. If v is the f.s.d. voltage across the meter, V is the desired new f.s.d. and Rm is the meter resistance, then:

v

vVRR m

FP3, questions 4 & 6b(use formula and picture on handout)

You will need to think a little for 6b.

PotentiometerA potentiometer is a variable potential divider

Moveable contact

Moveable

contact

Can be depicted as follows

PotentiometerA potentiometer is a variable potential dividerWhat is the resistance between points A and M?Resistance between points A and B can be estimated as: Rtot=rLtot/A (1)

Resistance between points A and M can be estimated as: : R1=rL1/A (2)

Dividing equation (1) by equation (2) we derive:

Rtot/R1=Ltot/L1

Thus, R1=RtotL1/Ltot

Or R1 /Rtot =L1/Ltot - just a simple ratio idea

Potentiometer

R1

Rtot

Ltot

A

B

L1

M

PotentiometerA potentiometer is a variable potential divider

V1

Vin

Rtot

R1 in

Vout

Potential divider

Potentiometer

R1

RtotR1=RtotL1/Ltot ; R1/Rtot=L1/Ltot

I=Vin/Rtot ; V1=IR1=VinR1/Rtot

Rto

t=R

1+

R2

V1=VinL1/Ltot

PotentiometerA potentiometer is a variable potential divider.

In order to change output voltage V1 we just need to shift a movable contact

V1

Vin

Rtot

R1

A potentiometer is a variable potential divider which may be used to measure an e.m.f.To measure e.m.f., it is necessary to draw no current though the unknown battery so that there is no voltage loss in the internal resistance.This can be done using a known driver cell, a potentiometer and a galvanometer.

Potentiometer

? V

1.5V

PotentiometerA potentiometer is a variable potential divider which may be used to measure an e.m.f.To measure e.m.f., it is necessary to draw no current so that there is no voltage loss in the internal resistance.This can be done using a known driver cell, a potentiometer and a galvanometer.The potentiometer is varied until the galvanometer reads zero (null detection) – the e.m.f. of the unknown can then be determined as a fraction of the known e.m.f.

1.5V

L1Ltot

e.m.f.= V1 =

VinL1/Ltot

V1

Vin

e.m.f.

PotentiometerA potentiometer is a variable potential divider which may be used to measure an e.m.f.To measure e.m.f., it is necessary to draw no current so that there is no voltage loss in the internal resistance.This can be done using a known driver cell, a potentiometer and a galvanometer.The potentiometer is varied until the galvanometer reads zero (null detection) – the e.m.f. of the unknown can then be determined as a fraction of the known e.m.f.

Wheatstone BridgePotentiometer rearranged to compare two resistances.Wheatstone bridge is a null detection circuit.

1.5V

I0=0

V0=0

Unknown R 3

Known R 4

R1 R2

Wheatstone BridgeThe galvanometer reads zero when V2=V4 and V3=V1 This can be seen from the second Kirchoff’s law for red and blue loops and the condition that p.d. between points A and B is zero (the galvanometer reads zero): V2+VAB=V4

V1+VAB=V3 ; VAB=0, thus: V2=V4 and V3=V1

From the first Kirchoff’s law for node A:

I2=I1+IAB with the current IAB between points A and B, but it is zero according to the galvanometer reading. Thus, I2=I1=IL where IL is the current flowing through the left branch of the bridge.

A B

I2

I1

Wheatstone BridgeFrom the first Kirchoff’s law for node B:

I4+IAB=I3 with IAB=0, thus, I4=I3=IR

Where IR is the current flowing through the right branch of the bridge

A B

IL

IL I3

I4

Wheatstone BridgeFrom the first Kirchoff’s law for node B:

I4+IAB=I3 with IAB=0, thus, I4=I3=IR

Where IR is the current flowing through the right branch of the bridge

From Ohm’s Law for all for resistors:

V1=I1R1=ILR1, V2=I2R2=ILR2, V3=I3R3=IRR3, V4=I4R4=IRR4

From equations V2=V4 and V3=V1 we derive:

IRR3 =V3=V1=ILR1

IRR4 =V4=V2=ILR1

IRR3 =ILR1, thus, IR =ILR1/R3

Substituting IR =ILR1/R3 into IRR4

=ILR1 we derive ILR1/R3 x R4 =ILR1 then dividing by

R4

Then dividing by R4 we obtain:

A B

IL

IL IR

IR

Wheatstone Bridge

A B

IL

IL I3

I4

Therefore R1/R2=R3/R4.R3 is usually unknown resistance, other three are known. In order to balance the bridge, the ratio R1/R2 is varied until galvanometer reads zero.Then R3=R4(R1/R2).

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kn

ow

n

Wheatstone BridgePotentiometer rearranged to compare two resistances.Unknown e.m.f. replaced by a divider that produces a fraction of the known p.d. Wheatstone bridge is a null detection circuit.Reads zero when V2=V4, V1=V3. Therefore R1/R2=R3/R4.R3 is usually unknown resistance, other three are known. Ratio R1/R2 is varied until the bridge balances, and galvanometer reads zero.Then R3=R1R4/R2.

The Wheatstone bridge measures resistance.Runknown=R1R4/R2

The Wheatstone bridge measures resistance.Runknown=R1R4/R2

=

This is just simple ratiosWade through presentation 5 if you would like to see the theory

Do Q7

Try the other questions on FP3

Next weekCapacitors

Bringing it all together

Revision and exam style practice questions

Friday – test. (NOT part of your final mark!)

If you want to do well…Study!

Textbook is very good

Ppts are on Learnzone