TAP 117- 2: Kirchhoff's laws - science-spark.co.uk€¦ · Web view · 2015-04-24I2R = E2R/(R +...

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Electrons, Photons and Waves Module 2.3: DC Circuits

©2011 science-spark.co.uk RAB Plymstock School

Module 2.3 DC Circuits

1. Kirchoff’s Laws 2.3.1 Series and parallel circuitsCandidates should be able to:(a) state Kirchhoff’s second law and appreciate that this is a consequence of conservation of energy;(b) apply Kirchhoff’s first and second laws to circuits;

conservatione.m.f.p.d.

2. Resistors in parallel and series

(c) select and use the equation for the total resistance of two or more resistors in series;(d) recall and use the equation for the total resistance of two or more resistors in parallel;

Resistance, Ohm, PD, Current, series, parallel

3. Internal resistance (e) solve circuit problems involving series and parallel circuits with one or more sources of e.m.f.;(f) explain that all sources of e.m.f. have an internal resistance;(g) explain the meaning of the term terminal p.d.;(h) select and use the equations e.m.f. = I (R + r), and e.m.f. = V + Ir .

Series, parallel, emf, od, internal resistance, power supply

4. Potential Dividers 2.3.2 Practical circuitsCandidates should be able to:(a) draw a simple potential divider circuit;(b) explain how a potential divider circuit can be used to produce a variable p.d.;(c)recall and use the potential divider equation Vout = Vin (R1 / R1 +

Circuit, Potential Difference, Potential Divider, Resistance

5. Potential Dividers (d) describe how the resistance of a light dependent resistor (LDR) depends on the intensity of light;(e) describe and explain the use of thermistors and light-dependent resistors in potential divider circuits;(f) describe the advantages of using dataloggers to monitor physical changes (HSW 3).

Circuit, Potential Difference, Potential Divider, Resistance

6. G482 Module 1: 2.3 DC Circuits Test

Review and assess their knowledge and understanding


Lesson 15 notes - Kirchhoff's laws

Objectives(a) state Kirchhoff’s second law and appreciate that this is a consequence of conservation of energy;(b) apply Kirchhoff’s first and second laws to circuits;

Kirchhoff's two laws are equations based on conservation of charge and conservation of energy that can be applied to any electric circuit. The two laws can be used to work out the current in any branch of a circuit, given the emfs and resistances in the circuit.

Kirchhoff's first law states that the total current entering a junction is equal to the total current leaving the junction.

Kirchhoff's second law states that the sum of the emfs round any complete loop in a circuit is equal to the sum of the potential drops round the loop.

Kirchhoff's first law is a statement of conservation of charge since it means that the total charge flowing into a junction in a given time is equal to the total charge leaving the junction in the same time. Kirchhoff’s second law is a statement of conservation of energy since an emf is where energy is supplied to charges (i.e. a source) and a potential drop is where charges release energy (i.e. a sink). The sum of the e.m.f s is therefore the total energy produced in the loop per unit charge and the sum of the potential drops is the total energy dissipated in the loop per unit charge.

Kirchhoff's first lawThe total current into a junction = the total current out of the junction.

Using the convention that currents leaving a junction are the opposite sign to currents entering the junction, the first law may be expressed as the following equation:

I1 + I2 + I3 + … = 0 where I1, I2, I3 etc represent the currents in the branches connected to the junction.


I1

I2

I3

I4

I1 + I2 + I3 + I4 = 0

currents in are assigned + valuescurrents out are assigned – values

Kirchhoff’s 1st law


Kirchhoff's second lawThe sum of the emfs round a loop in a circuit = the sum of the potential drops round the loop.

For a loop of a circuit containing emfs 1, 2, 3, etc and resistances R1, R2, R3, etc, the second law may be expressed as the following equation:

1 + 2 + 3 + … = I1R1 + I2R2 + I3R3 + … where I1, I2, I3 … represent the currents through the resistances R1, R2, R3 ….

I1 – I2A I1

r1

B C

D

I1

I2

I1 – I2

= I2R + I1r1 for loop ABCD

Using Kirchhoff’s second law

I2

R

Notes:1. emfs and currents in the opposite direction to the direction round the loop are given negative values.

2. The second law is particularly useful for analysing circuits with more than one loop. In general, for a circuit with n unknown currents, n loops need to be considered one by one, each loop giving a new equation. In this way, the n linear equations formed can be solved for the n unknown currents.

Example:A circuit consists of a cell of emf 1.6 V in series with a resistance 2.0 connected to a resistor of resistance 3.0 in parallel with a resistor of resistance 6.0 Determine the total current drawn from the cell and the potential difference across the 3.0 resistor.


I1 + I2

I1

I2

1.6 V2.0

3.0

6.0

I1 + I2

SolutionConsider the circuit loop consisting of the cell and the 3.0 resistor:

1.6 V = 3 I1 + 2 (I1 + I2).

Thus: 1.6 V = 5 I1 + 2 I2.

Consider the circuit loop consisting of the cell and the 6.0 resistor:

1.6 V = 6 I2 + 2 (I1 + I2).

Thus 1.6 V = 2 I1 + 8 I2.

Subtracting the second equation from the first gives:

0 V = 3 I1 + 6 I2

hence I1 = 2 I2.

Substituting I1 = 2 I2 into the second equation gives:

1.6 V = 12 I2.

Thus I2 = 0.13 A and I1 = 0.27 A.

Current through cell = I1 + I2 = 0.40 A.

pd across 3.0 resistor= I1 × 3.0 (= I2 6.0 ) = 0.8 V.


Lesson 15 Questions – Kirchoff’s Laws1)a) State Kirchoff’s Second Law………………………………………………………………………………………………………....……………………………………………………………………………………………………………………………………………………………………………… (2)

Fig.1.1 shows a circuit:

fig 1.1b) Use Kirchoff’s second law to determine the p.d. V.………………………………………………………………………………………………………....……………………………………………………………………………………………………………………………………………………………………………… (2)

Total [4]

2)a) Kirchoff’s first law is based on the conservation of an electrical quantity. State the law and the quantity conserved.………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

b) Fig 2.1 shows a potential divider circuit.


fig 2.1


i) Show that the voltmeter reading is 1.5V…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

Total [4]

3) Fig 3.1 shows part of an electric circuit.

fig 3.1a) Name the component marked X………………………………………………………………………………………… (1)

b) Determine the magnitude of the currents I1, I2 and I3.I1 =………………………mAI2 =………………………mAI3 =………………………mA

(3)Total [4]


Lesson 16 notes - Resistor Combination Formulae

ObjectivesBe able to:Recall and use formula for the combined resistance of two or more resistors in seriesRecall and use formula for the combined resistance of two or more resistors in parallel

Resistors in combinationsRemember that R = V / I where V is the pd across a component and I is the current through it.

Resistors in series

From Kirchoff’s first law:Current constant around series circuit

From Kirchoff’s second law:E=V1 + V2and VT = V1 + V2Therefore IRT = IR1 + IR2Cancelling I


VVTTAA

VV11

VV22

II

RR11 RR22

EE

RT = R1 + R2

This is true for any number of resistors in series.

Resistors in parallel

From Kirchoff’s first law:

I1 = I2 + I3 (1)

Applying Kirchoff’s scond law to loop 2:V2 + V1 = E0So, (I3 x R2) – (I2 x R1) = 0 VWhich leads to I3 x R2 = I2 x R1

This is saying that the 2 resistors have the same p.d. across them.So we can write,I1 = V/RTI2 = V/R1I3 = V/R2


RR22

RR11

EE

II22

II11

II33

Loop Loop 11

Loop Loop 22

Enter these into (1):

V/RT=V/R1 + V/R2

Cancel V

1/RT = 1/R1 + 1/R2

This is true for any number of resistors in parallel.

So RT is the reciprocal of 1/RT

In AS level physics most problems encountered involve 2 resistors in parallel and for this we can write.

RT = R1R2/(R1+R2)

Because1/RT = 1/R1 + 1/R2

and so, 1/RT = R2/(R1xR2) + R1/(R2xR1)1/RT = (R2 + R1)/(R2xR1)

Therefore, RT = (R2xR1)/(R2 + R1) or RT = R1R2/(R1+R2)


Lesson 16 questions – resistors in series and parallel

1 Calculate the total resistance of the following resistance combinations:

a)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (1)b)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)c)

…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (3)

Total [6]


3Ω

6Ω

2Ω

4Ω

12Ω

4Ω 12Ω

2) Calculate the total resistance between the points X and Y for the circuit shown in fig. 1.1

fig1.1……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (3)

Total [3]

3) A circuit of a combination of resistors is shown in figure 3.1.

fig 3.1Calculate:a) the combined resistance of the 3Ω resistor and the 6Ω resistor in parallel,…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)b) the total resistance of the circuit,…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (1)c) the battery current,…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)d) the power supplied to the 4Ω resistor.…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (3)

Total [8]


3Ω

6Ω

4Ω

6V

Lesson 17 notes – Internal resistance

ObjectivesYou will be able to:Use the concept that E.M.F. is defined in terms of the energy transferred by a source in driving unit charge round a complete circuit.Use energy considerations to distinguish between E.M.F. and P.D. Appreciate that sources of E.M.F. have internal resistance and understand the simple consequences of internal resistance for external circuits.

What you should know

E.M.F is the gaining of energy by a charge as it passes through a supply

Potential difference is the transferring of energy from the charge to a resistor

Ohm’s Law (V=IR) Voltage = Energy / Charge Current = Charge moved / time taken Kirchoff’s 1st Law (KCL) Kirchoff’s 2nd Law (KVL) Power = current x voltage (also P=I2R)

Internal ResistanceCurrent flows through all parts of the circuit, including the cells. Since the cells themselves are made of material with resistance there will be some energy lost in the cell. This will increase with current, so there is more loss at higher currents. Consider the diagram below:

Here we represent a ‘real’ cell as two components in a circuit diagram:

an ‘ideal’ cell, with no internal resistance, and a small resistor, representing the internal resistance.


Electrical energy to heat: Internal resistance r

Chemical energy to electrical energy: Cell of emf E

Electrical energy to work and heat: external load resistance R

CELL

Energy is conserved, so the energy transferred from chemical to electrical by the emf of the cell must ‘pay’ for the energy transferred to other forms in the circuit - this is the sum of the energy transferred in the cell itself and the energy transferred by the external load.

Clearly the energy transferred to heat in the cell is not doing useful work and so is in a sense ‘lost’.

Deriving an equationThere are three ways to arrive at the equation relating emf, terminal pd, current and internal resistance.

1. Kirchhoff’s 2nd Law: As charge goes around the circuit the sum of e.m.f s must equal the sum of voltage drops leading to:

E = I R + I rThe terminal voltage is equal to I R so this can be rearranged to give:

V = E – I rand interpreted as terminal voltage = emf – ‘lost volts’

2. Energy is conserved. Imagine a unit of charge, Q, moving around the circuit:

Q E = Q I R + Q I rThis leads to the same equations as in (1) above.

3. Use Ohm’s law with E ‘driving’ current through the combined resistance (R + r):

I = E / (R+r)Multiplying throughout by (R+r) leads to the same equations and conclusions as in (1).


CELL

Terminal Voltage

e.m.f‘lost volts’

R

rE

I

Practical effects of internal resistanceLets us a car as an example. The headlamps are connected in parallel across a twelve-volt battery. The starter motor is also in parallel controlled by the ignition switch. Since the starter motor has a low resistance it demands a very high current (say 60 A). The battery itself has a low internal resistance (say 0.01 Ω). The headlamps themselves draw a much lower current.

What happens when the engine is started? (switch to starter motor closed for a short time).

sudden demand for more current

large lost volts (around 0.01 Ω 60 A = 6 V)

terminal voltage drops to 12 V – 6 V = 6 V

headlamps dim

When the engine fires, the starter motor switch is opened and the current drops. The terminal voltage rises and the headlamps return to normal. It’s better to turn the headlamps off when starting the car.

Voltage, Current and Resistance

The voltage of a supply decreases if it is required to supply more current across the same load (R=V/I)The cell p.d. decreases as the current increases. (the internal resistance stays constant)The graph is shown opposite.E = IR + Ir becomes IR = E – IrCompare this to y = c +mxThe intercept with the y axis gives the supply e.m.f The intercept with the x axis gives the maximum current that can be drawn from the supplyAnd the gradient gives the internal resistance of the supply


V

I

Power and internal resistanceElectrical power, P = IV = I2R So power supplied by the cell,

IE = I2R + I2r

Output Power = power delivered to R + power wasted in the cell due to its internal resistance

The power supplied to R = I2R = E2R/(R + r) 2

And because of this the peak of the power verses resistance curve is at R = r, Maximum power is delivered when the load resistance is equal to the internal resistance. (When the load is matched to the source)

The graph plotting the Output power against Load resistance for the internal resistance of 1Ω shows this the best. As the load approaches 1Ω, the output power is at a maximum.


Lesson 17 questions – Internal resistance1) Fig. 1.1 shows two soldered resistors connected across terminals of a cell.

fig 1.1The cell has a resistance of 0.8Ω and e.m.f. 1.5V.a) Define e.m.f. in terms of energy transformed and electric charge.…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

b) Suggest why a cell has internal resistance.………………………………………………………………………………………… (1)

c) Calculate the total resistance R of the circuit in Fig. 1.1.……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

d) Hence calculate the current I in the circuit.……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

e)i) Write an equation for the power dissipated by a current-carrying resistor.……………………………………………………………………………………………………………………………………………………………………………………

ii) For the circuit in Fig. 1.1, calculate the ratio:

power dissipated by internal resistance__ power dissipated by total external resistance

ratio=……………(4)Total [7]

2) Fig. 2.1 shows a cell of e.m.f. E and internal resistance r connected to a variable resistor.


fig 2.1 fig 2.2

Fig. 2.2 shows the variation of the p.d. V across the terminals of the cell with the current I drawn from the cell.a) Explain how fig 2.2 shows that the e.m.f. E is 1.4V………………………………………………………………………………………………………………………………………………………………………………… (1)

b)i) Use fig. 2.2 to determine the maximum possible current that can be drawn from the cell.………………………………………………………………………………………………………………………………………………………………………………… (1)

ii) Calculate the internal resistance r of the cell.……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

Total [5]3) Fig. 3.1 shows the I/V characteristics for a tungsten filament lamp.

fig 3.11A 5.0Ω resistor and the tungsten filament lamp are connected in series to a d.c. power supply of e.m.f. 24V. The current drawn from the power supply is 2.0 A.


a) Calculate the total power delivered by the supply.……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

b) Use fig 3.1 to determine the resistance of the filament lamp when the current in it is 2.0 A.…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)c) Calculate the total resistance of the series combination of the filament lamp and the resistor.…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (1)d) Calculate the internal resistance of the supply.…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

Total [7]4) Fig 4.1 shows a car battery of e.m.f 12V and internal resistance 0.014Ω connected to the starter motor of a car. When the car engine is being started, the car battery provides a currnt of 160A to the starter motor.

fig 4.1a) Show that the p.d. across the internal resistance is about 2.2V.…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (1)

b) Determine the terminal p.d. across the battery.…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (1)

Total [2]


Lesson 18 notes – Potential dividersObjectivesYou will understand the use of a potential divider as a source of variable p.d.You will be able to describe and explain the use of thermistors and light-dependent resistors in potential dividers to provide a potential difference which is dependent on temperature and on light intensity respectively.

Thermistor Characteristics

Ohmic-Conductor keeping temperature constant

Potential DividersPotential dividers are resistors connected in series across a voltage source used to obtain a desired fraction of the voltage. An example is shown below:


Low temp

Increased temp

The potential divider formulaFor an unloaded potential divider the current through each resistor is the same so the voltage is proportional to the resistance. This means that the pd across the pair of resistors is divided in the same ratio as the resistors themselves:

i.e. V1 / V2 = I R1 / I R2 or V1 / V2 = R1 / R2

If R1 >> R2 then V1 is more or less the supply voltage and if R1 << R2 then V1 is close to 0 V. VS is an input to the potential divider and V1 is an output. The circuit itself provides a way to tap off a voltage between 0 V and VS.

This can, of course be done continuously using a rheostat or potentiometer shown below:

Rotary potentiometers are used as volume controls in hi-fi systems.

The potential divider equation can be derived by rearranging the ratios above to give:

V1 = VS x (R1 / (R1+R2))

Lets do it now:

Use KVL: Vs = I(R1+R2) so I = Vs/(R1+R2)

Now: V1/V2 = IR1/IR2 = R1/R2

And V1=I x R1

So, V1= (Vs/(R1+R2)) x R1

Therefore: V1 = VS x (R1 / (R1+R2))


ExtensionThe effect of the load on outputConnecting a load across R1 reduces the output voltage.

This is because the effective resistance in the lower arm of the potential divider is now a parallel combination of R1 and R load (less than R1) so a smaller fraction of the voltage is ‘tapped off’.

If Rload >> R1 then there is no significant effect on the output voltage.

Consider what happens when a lit bulb goes out when “shorted out” by a piece of wire. It is because the low resistance of the wire in parallel reduced the combination’s total resistance, compared to the rest of the circuit.


Lesson 14 questions – Potential dividers1) A potential divider circuit based on a light-dependent resistor (LDR) is shown in fig 1.1. The supply has negligible internal resistance.

fig 1.1a) The light intensity falling upon the LDR is increased. State:i) how the resistance of the LDR changes,…………………………………………………………………………………………….……………………………………………………………………………………… (1) ii) how the p.d. across the LDR changes.…………………………………………………………………………………………….……………………………………………………………………………………… (1)

b) At a particular light intensity, the resistance of the LDR is 420Ω.i) Calculate the p.d. across the LDR.…………………………………………………………………………………………….…………………………………………………………………………………………….…………………………………………………………………………………………….…………………………………………………………………………… (3)ii) An electronic circuit of resistance 50Ω is connected between the terminals X and Y.

1. Show that the total resistance of the parallel combination of this electronic circuit and the LDR is about 45Ω.…………………………………………………………………………………………….…………………………………………………………………………………………….…………………………………………………………………………………………….…………………………………………………………………………… (1)

2. Calculate the p.d. across the electronic circuit.…………………………………………………………………………………………….…………………………………………………………………………………………….…………………………………………………………………………………………….…………………………………………………………………………… (2)


Total [8]


2) Fig 2.1 shows a potential divider circuit. The battery has negligible internal resistance and the voltmeter has a very high resistance.

fig2.1i) Show that the voltmeter reads 1.5V.…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

ii) An electric device rated at 1.5V, 0.1A is connected between the terminals X and Y. The device has constant resistance. The voltmeter reading drops to a very low value and the device fails to operate, even though the device itself is not faulty.

1. Calculate the total resistance of the device and the 400Ω resistor in parallel.……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (3) 2. Calculate the p.d. across the device when it is connected between X and Y.……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

3. Why does the device fail to operate?……………………………………………………………………………………… (1)

Total [8]3)a) Fig 3.1 shows the circuit symbol for a component.

fig 3.1Name the component and state how its resistance changes as the temperature of the component is increased.………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)


b) Fig 3.2 shows a potential-divider circuit. The battery has a negligible internal resistance and the voltmeter has a very high resistance.

fig 3.2i) At a particular temperature, the resistance of component X is 4.2kΩ. Calculate the voltmeter reading.………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (3)

ii) State how your answer to (b)(ii) changes when the temperature of the component X is increased.…………………………………………………………………………………………………………………………………………………………………………………… (1)

Total [6]


Lesson 18 notes – Potential dividersObjectives(a) draw a simple potential divider circuit;(b) explain how a potential divider circuit can be used to produce a variable p.d.;(c)recall and use the potential divider equation Vout = Vin (R1 / R1 + R2 ).

Thermistor Characteristics

Ohmic-Conductor keeping temperature constant

Potential DividersPotential dividers are resistors connected in series across a voltage source used to obtain a desired fraction of the voltage. An example is shown below:


Low temp

Increased temp

The potential divider formulaFor an unloaded potential divider the current through each resistor is the same so the voltage is proportional to the resistance. This means that the pd across the pair of resistors is divided in the same ratio as the resistors themselves:

i.e. V1 / V2 = I R1 / I R2 or V1 / V2 = R1 / R2

If R1 >> R2 then V1 is more or less the supply voltage and if R1 << R2 then V1 is close to 0 V. VS is an input to the potential divider and V1 is an output. The circuit itself provides a way to tap off a voltage between 0 V and VS.

This can, of course be done continuously using a rheostat or potentiometer shown below:

Rotary potentiometers are used as volume controls in hi-fi systems.

The potential divider equation can be derived by rearranging the ratios above to give:

V1 = VS x (R1 / (R1+R2))

Lets do it now:

Use KVL: Vs = I(R1+R2) so I = Vs/(R1+R2)

Now: V1/V2 = IR1/IR2 = R1/R2

And V1=I x R1

So, V1= (Vs/(R1+R2)) x R1

Therefore: V1 = VS x (R1 / (R1+R2))


ExtensionThe effect of the load on outputConnecting a load across R1 reduces the output voltage.

This is because the effective resistance in the lower arm of the potential divider is now a parallel combination of R1 and R load (less than R1) so a smaller fraction of the voltage is ‘tapped off’.

If Rload >> R1 then there is no significant effect on the output voltage.

Consider what happens when a lit bulb goes out when “shorted out” by a piece of wire. It is because the low resistance of the wire in parallel reduced the combination’s total resistance, compared to the rest of the circuit.


Lesson 18 questions – Potential dividers1) A potential divider circuit based on a light-dependent resistor (LDR) is shown in fig 1.1. The supply has negligible internal resistance.

fig 1.1a) The light intensity falling upon the LDR is increased. State:i) how the resistance of the LDR changes,…………………………………………………………………………………………….………………………………………………………………………………… (1) ii) how the p.d. across the LDR changes.…………………………………………………………………………………………….………………………………………………………………………………… (1)

b) At a particular light intensity, the resistance of the LDR is 420Ω.i) Calculate the p.d. across the LDR.…………………………………………………………………………………………….…………………………………………………………………………………………….…………………………………………………………………………………………….…………………………………………………………………………… (3)ii) An electronic circuit of resistance 50Ω is connected between the terminals X and Y.

1. Show that the total resistance of the parallel combination of this electronic circuit and the LDR is about 45Ω.…………………………………………………………………………………………….…………………………………………………………………………………………….…………………………………………………………………………………………….……………………………………………………………………………… (1)

2. Calculate the p.d. across the electronic circuit.…………………………………………………………………………………………….…………………………………………………………………………………………….…………………………………………………………………………………………….…………………………………………………………………………… (2)

Total [8]


2) Fig 2.1 shows a potential divider circuit. The battery has negligible internal resistance and the voltmeter has a very high resistance.

fig2.1i) Show that the voltmeter reads 1.5V.…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

ii) An electric device rated at 1.5V, 0.1A is connected between the terminals X and Y. The device has constant resistance. The voltmeter reading drops to a very low value and the device fails to operate, even though the device itself is not faulty.

1. Calculate the total resistance of the device and the 400Ω resistor in parallel.………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (3)

2. Calculate the p.d. across the device when it is connected between X and Y.………………………………………………………………………………………………………………………………………………………………………………… (2)

3. Why does the device fail to operate?……………………………………………………………………………………… (1)

Total [8]3)a) Fig 3.1 shows the circuit symbol for a component.

fig 3.1Name the component and state how its resistance changes as the temperature of the component is increased.……………………………………………………………………………………………………………………………………………………………………………………………… (2)


b) Fig 3.2 shows a potential-divider circuit. The battery has a negligible internal resistance and the voltmeter has a very high resistance.

fig 3.2i) At a particular temperature, the resistance of component X is 4.2kΩ. Calculate the voltmeter reading.………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………(3)

ii) State how your answer to (b)(ii) changes when the temperature of the component X is increased.…………………………………………………………………………………………………………………………………………………………………………………(1)

Total [6]


TAP 117- 2: Kirchhoff's laws - science-spark.co.uk€¦ · Web view · 2015-04-24I2R = E2R/(R +...

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Transcript of TAP 117- 2: Kirchhoff's laws - science-spark.co.uk€¦ · Web view · 2015-04-24I2R = E2R/(R +...