# Concepts of Electrical Principles

**Disclaimer:** This work has been submitted by a student. This is not an example of the work written by our professional academic writers. You can view samples of our professional work here.

Any opinions, findings, conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of UK Essays.

Published: *Fri, 18 May 2018*

Everything is made of atoms; in turn atoms consist of a combination of minuscule particles known as neutrons, protons and electrons. The nucleus of an atom consists of protons and neutrons while electrons exist in a cloud surrounding and rotating around the nucleus. The electron and proton are capable of holding an electrical charge; electrons hold negative charges and protons positive charge. We know that like charges repel each other while opposite charges have the opposite effect in attracting one another.

If we wish to measure the flow of electrons around a circuit we refer to this as a measure of electrical current. Electric current is represented by the symbol I and is a quantity of charge carriers passing a given point in a circuit. This is calculated as coulomb of charge passing a defined point in one second, which as a unit is given the name ampere abbreviated to A. This can be measured using an instrument called an ammeter which when connected in series with a circuit to measure the current passing through it.

For electric current to flow around a circuit there must be a voltage across it. Voltage is a measure of the potential difference (p.d), which acts like electric pressure pushing the current around the circuit. The pressure can be read in a circuit by a voltmeter, which must be applied through the resistance. This happens when there is a deficit of electrons in a conductive material and this is then connected to another material with excess electrons. This is the case in a battery where chemicals allow electrons to flow from the negative terminal that contains an excess of electrons and the positive terminal containing positively charged protons. This happens because opposite charges attract one another.

#### 1.4 Resistance

This flow of current faces opposition from resistance; this is a quantity of how much the electrons bump against the particular conductor they are flowing through. Some materials conduct electricity better then others. Materials that have a high resistance conduct electricity less well. Resistance limits the flow of electrons between the positive and negative ends of a circuit. We measure resistance in units called ohms (Î©). One ohm is defined as the amount of resistance you have in a conductor when applying one volt of electrical pressure creates one amp of current.

#### 1.5 Energy

When electrons sit high in there shells surrounding the nucleus they have electrical energy. This energy can be harnessed to do work in various ways, if the electrons bump into atoms this can cause them to move around which creates heat, they create electromagnetic waves as they travel which can use there attraction and repulsion to move things magnetically, and if the electrons move down there electric shells they give up excess energy giving out light in the manner of photons.

#### 1.6 Charge Carriers

The sub-atomic particles that carry charge are known as protons and electrons as previously discussed electrons are negatively charges while protons are positively charged. The unit to measure the quantity of electrical charge (Q) is the coulomb (C) where 1 coulomb of charge is equal to charged electrons. If one coulomb of charge passes a point in one second we say this is one ampere of current. We can use our knowledge of math to deduct that if then if we take (I) as the current in amperes and t as the time in seconds then:

Electrical Principles/

### Kirchhoff’s Laws

#### 2.1 Potential Difference

The pull created by the difference in charge between the two sides of a circuit is called the potential difference, which is otherwise known as the voltage. Voltage sources that have higher attractive forces are known to have a higher potential difference.

The units we use to measure voltage/potential difference is known as the ampere which is explained in section 1.6 as one coulomb of charge passing a given point in one second.

#### 2.2 Ohms Law

a) Ohms law relates Voltage, Current and Resistance in the following equation:

I = current in amperes

V = voltage in volts

R = Resistance in Ohms

This law states that the current I flowing in a circuit is directly proportional to the voltage applied to it and inversely proportional to the resistance.

b) For a 5m length of wire with a resistance of 600 ohms we can apply this law. If you where to half the length of wire you would half the resistance as there would be half as much material for the electrons to bump into.

c) If we where to increase the length of the wire to 8m we can see that the resistance increases as create more material for the electrons to crash into.

d) To find the length of the same wire when the resistance is 420 ohms we do the following sum:

So we can say that the same wire with a resistance of 420ohms would measure 3.5 meters.

#### 2.3 Resistance Variation

If a piece of wire has a cross sectional area of 2mm2 and a resistance of 300 ohms

Find the resistance of the same length of wire if the cross sectional area is 5mm2.

Given that resistance is inversely proportional to cross sectional area, increasing the cross sectional area increases the flow of electrons, we can calculate this mathematically as such:

b) Find the cross sectional area of a wire of the same length and material of resistance 750.

#### 2.04

Calculate the resistance of a 2km length of aluminium overhead power cable if the cross sectional area of the cable is 100mm2. Take the resistivity of aluminium to be 0.03 x 10-6 â„¦m

Show the equation you are using in your answer.

We know that and that if we combine these rules we can create the formula . With one more piece of information we will be able to take the material used into account.

This is done by including the resistivity of the material into the relationship treating it as a constant of proportionality. We use the symbol Ï (Greek rho). The final equation will look like this:

#### 2.5 Power

If electrical energy (W) = Charge (Q) x Voltage (V) then :-

a) Show the equation for power in terms of current( I) and voltage (V).

Electrical Energy (W) = Charge (Q) x Voltage (V)

W = Q x V

Power (P) = Current (I) x Voltage (V)

P= V x I

b) Using Ohms law explain how power can also be expressed in terms of I and R, and, V and R.

P= V2/R

P = I2R

C) An e.m.f. of 250V is connected across a circuit resistance and the electric current through the circuit resistance is 4A. What is the power dissipated in the circuit?

#### 2.6

a) To discover the potential difference across the winding we use Ohms law as follows:

Voltage (V) = Current (I) X Resistance (R)

V= 5A X 100Î©

V = 500V

b) If we wish to find the power dissipated by that same coil we use our equations for power:

Power (W) = Voltage (V) X Current (I)

P= 500V x 5A

P = 2500 Watts

#### 2.7

A 12V battery is connected a load having a resistance of 40Î©.

a) Determine the current flowing in the load.

For this we must again use Ohms law rearranged to make I the subject.

I = V /R

I = 12V / 40Î©

I = 0.3 Amps

Determine the power consumed by the load.

To calculate this we use our power equation again using the figure we just calculated for the current.

P = VI

P = 12V x 0.3A

P = 3.6 watts

c) Determine the electrical energy dissipated in 2 minutes.

Electrical Energy (W) = Charge (Q) x Volts (V)

Current is charge per second and we discovered that this circuit runs 0.3Amps, finding how much energy is dissipated in 2mins first requires changing minutes to seconds.

2mins = 120 seconds

W = Q x V

W = (120 X 0.3) x 12V

W = 432 Watts

#### 2.8

a) Explain what is meant by one unit of electricity with reference to Electrical Charge (Q), Voltage (V) and Time (T).

A standard unit of electricity is usually calculated as a Kilowatt-hour (KWh), Which is 1000 watts of electricity dissipated for one hour. SEE MY BOOK ON THIS

b) Determine the power dissipated by the element of an electric fire of resistance 20â„¦ when a current of 10A flows through it.

For this situation we are provided with the current at 10A and the resistance at 20â„¦ therefore we can use our power equation to find how much power is dissipated.

P = I2R

P = 102 x 20

P = 2000 watts

c) If the fire is on for 6 hours determine the energy used and the cost if 1 unit of electricity costs 13p.

Firstly we take the power consumption in watts from we determined in question b then apply the following equation to it:

Cost per Unit x Watts / 1000

Multiply the per-hour cost by the running time.

26p x 6h = £1.56p

#### 2.9

Analyse this resistors in series’ circuit:

a) Express ‘V’ in terms of V1, V2 and V3.

VT = V1 + V2 + V3

Voltages in this circuit will each have a different value if the resistances are different but if you add all the values together they should in total equal the supply voltage.

b) Express the total circuit resistance (RT) in terms of R1, R2 and R3.

Resistances in series always add together.

This can be expressed as:

RT = R1 + R2 + R3

c) Express in terms of ‘I’ what the electric current is through the ammeter-A, R1, R2 and R3.

In a series circuit the current is the same in any part of the circuit so readings using the ammeter would be the same as any reading taken on each of the resisters R1, R2 or R3.

#### 2.10

A 12V battery is connected across a circuit having three series-connected resistors of resistances 4â„¦, 9â„¦ and 11â„¦.

a) Determine the electric current through the circuit.

As this is a series circuit the current would be the same throughout the circuit, to calculate this we must use ohms law, first we know that resistances add together in a series circuit to give the resistance total.

4â„¦ + 9â„¦ + 11â„¦ = RT = 24â„¦

Then we must implement Ohms law:

I =V/R

I = 12V / 24â„¦

I = 0.5A

b) Determine the p.d. across the 9â„¦ resistor.

Via Ohms law and our previous current calculation, we calculate the voltage across the 9â„¦ resister.

V2 = I x R1

V2 = 0.5 x 9

V2 = 4.5 Volts

c) Determine the power dissipated in the 11â„¦ resistor.

P = I2R

P3 = 0.52 x 11

P3 = 2.75 W

#### 2.11

Two resistors are connected in series across a 24V supply with a flow of electric current of 3A within the circuit. If one of the resistors has a resistance of 2â„¦ determine:

a) The value of the other resistor.

R2 = RT – R1

R2 = 8 – 2

R2 = 6â„¦

Trusting in Ohms law we can find the value of the other resistor using the values given for total voltage and current and knowing that resistances in series add together to give the resistance total.

RT = V/I

RT = 24/3

RT = 8â„¦

b) The p.d. across the 2â„¦ resistor.

Solving this requires Ohms law.

V1 = I x R1

V1 = 3A x 2Î©

V1 = 6 Volt

c) How much energy is used if the circuit is connected for 50 hours.

P=VI

P=24v x 3

P=72W

50h = 180000s

W = Q x V

Q (charge) = I (current) x t (time)

W =180000 x 72

W = 12960000 Watt/joules

#### 2.12

Analyse the ‘resistors in parallel’ circuit.

a) In terms of ‘V’, express the p.d. across R1, R2, and R3.

V= I1R1 = I2R2 = I3R3

We see that the voltage is the same across each resistor.

b) Express the total load current ‘I’ in terms of I1, I2, and I3.

#### 2.13

For the circuit shown below, determine:

a) The reading on the ammeter,

In a purely parallel circuit the voltage will be the same in each branch of the circuit.

V=I x R

V = I1 x R1

V = 8 x 5 = 40V

I = V/R3

I = 40/20 = 2A

b) The value of resistor R2.

We now have all the values for I,

= 11 – 8 – 2 = 1A

R2 = V/I2

R2 = 40/1

R2 = 40â„¦

#### 2.14

Find the value of resistor that can replace the six resistors in this diagram.

We know that resistances in series can be added together to give the total resistance, in this example we have a parallel network of resisters in series with 3 more resisters. Treating this parallel network as a single resistance will allow us to calculate the total resistance of the circuit easily.

Convert the resistances to conductance:

Adding them together gives us the total conductance

0.52G

This can then easily be converted to resistance.

Now the parallel circuit can be treated as a single resister, we can add all the resistors together and find the total resistance of the circuit giving us the value of a resister we can replace it with.

#### 2.15

Analyse the circuit below and determine:

The currents I1, I2, I3, I4, I5, and I6

We can treat the two sets of parallel resisters as single resisters if we first convert them to conductance and then for each add the conductances together then convert back to resistance.

For the set of 3 parallel resisters:

The Set of two:

The three resisters can be added to give our RT

We can now add these conductances together giving us our total conductance for the set of two resistors. This can then be converted to a combined resistance easily:

We now proceed to do this for the set of three resistors:

We now have the equivalent of 3 resistors in series, which we know can be added together to create a single resistance:

Now that we know the total resistance for the circuit we can find I1 easily using Ohms law:

We must now find the voltages V1, V2 and V3 in order to later find the currents through the network branches.

=20V

And now V2

Next I will calculate V3

We can check this by adding all of the voltages to see if they equal the total voltage we have been given.

This is over by 1.4V but I believe this is due to the compound effects of the rounding bug and that the calculations made are correct.

We know that the current through I1 is 5A now we will work out the currents through the branches of the parallel resistances using Ohms law:

#### 2.16

State Kirchoff’s first (current) law. Show that the currents I2 and I3 combined are equal to the input current I1

Kirchhoff’s Current Law states:

The sum of the currents entering a particular point must be zero.

So all currents entering a point must equal all the currents flowing from it. Therefore we must now think of the currents flowing from the junction as negative currents.

i1Â +Â i2Â +Â i3Â +Â i4Â = 0

Observing our circuit we see 11A of current going in, this means that the same amount of current must come out.

Therefore

To prove this we calculate I1 and I2 using Ohm’s law

I2= V/R

I2=10/10

I2= 1A

I3= V/R

I3=10/1

I3= 10A

We can now calculate I1 expecting it to equal our given figure of 11A.

I1= I2 + I3

I1=10+1

I1=11A

#### 2.17

Using Kirchhoff’s first (current) law, calculate current I1 and I2 in the network below.

Kirchhoff’s first current law states that the sum of the current entering a point must be zero.

Examining the junctions we have 1.2A and 4.5A flowing in and 0.6A and I1 are flowing out.

1.2 +4.5 = I1+0.6

1.2 + 4.5 – 0.6 = I1

I1= 5.1A

For I2 there are three currents flowing in but none flowing out. This must mean that the last value is a negative value.

5.1+3 + I3 = 0

8.1 + I3 =0

I3 = – 8.1A

#### 2.18

The potential divider shown below is used as a simple voltage calibrator. Determine the output voltage produced by the circuit:

(a) When the output terminals are left open-circuit (i.e. when no load is connected);

We can solve this using the Voltage Divider Rule.

Connecting a resistor to V-out will create a parallel resistor network. We can use the product over sum formula to find the comparable resistance because there are only two resistors.

to 1 dp

With this information we can calculate the voltage.

V=0.2V 1dp

#### 2.19

A moving coil meter requires a current of 1 mA to provide full-scale deflection. If the meter coil has a resistance of 100â„¦ and is to be used as a milliammeter reading 5 mA full-scale, determine the value of parallel shunt resistor required.

REVIEW ME

Make the meter useable over 5ma by adding a resistor to “switch” the range of the meter like you would on a none autorangeing multimeter.

This is done by adding a resistor IN PARALLEL with the meter.

#### 2.20

Two resistors, one of 15 â„¦ and one of 5 â„¦ are connected in parallel. If a current of 2 A is applied to the combination, determine the current flowing in each resistor.

As there is only two resistors we can use our product over sum equation to find the total value of resistance the parallel network provides.

Using this we are now able to find the voltage.

Now we can find the current through each branch,

I1:

I=V/R

I1 = 7.5/15

I1 = 0.5A

I2:

I=V/R

I2 = 7.5/5

I 2= 1.5A

#### 2.21

A switched attenuator comprises five 1 kâ„¦ resistors wired in series across a 5V d.c. supply.

If the output voltage is selected by means of a single-pole four-way switch, sketch a circuit and determine the voltage produced for each switch position

1KÎ©

1KÎ©

5V

Switch

1KÎ©

1KÎ©

Vout

Answer: 1V, 2V, 3V, 4V, 5V

#### 2.22

With the aid of a diagram, briefly explain in your own words Kirchhoff’s second law.

In an electronic loop the sum of all the voltages around the circuit taking polarity into account will equal zero.

For example if you where to travel around a circuit following conventional current taking the voltage at each resistance including the battery and added all of those voltages up including negative voltages the sum would equal zero. We would see that the battery would give the circuit charge a EMF while all of the resistances would dissipate this force.

#### 2.23

Using Kirchhoff’s second law, determine the value of e.m.f. (E) in the circuit below.

E+5=14

E= 14-5

E=9V

2.24

Using Kirchhoff’s laws together with the use of simultaneous equations, determine the current flowing in each branch of the network shown in the circuit below.

Here we are presented with essentially two loops of current where readings in the connecting part of the loops will be affected by one another. We will use Kirchoffs laws to solve the problem by first treating the current as two separate loops.

We use simultaneous equations to find our two unknowns I1 and I2 .

Loop Two

E2 = I2r2 + (I1 + I2)R

2 = I2 + 4I1 + 4I2

2 = 4I1 + 5I2

Loop One

E1 = I1r1 + (I1 + I2)R

4 = 2I1 + 4I1 + 4I2

4 = 6I1 + 4I2

6I1 = 4 – 4I2

Substitute I1 into the second loop.

Amps

As we have obtained I1 we can now work on I2

4 = 6I1 + 4I2

R=I1+I2

2.25

Analyse the circuit shown below and determine the following parameters

a) The current in each branch of the circuit.

I1 =

I2 = 1.233A

b) The voltage across the load resistance.

0.426

c) The power dissipated by the load resistor.

P=

d) Use computer software to verify your results.

26) A temperature sensor is connected into a bridge measuring circuit as shown. If the value of the sensor is 110R at 0oC and it increases by 0.2% for every degree the temperature rises and falls a corresponding amount if the temperature drops.

What voltage will be output on the voltmeter when the temperature is :-

(a) 25oC

(b) 100oC

(c) -40oC

Build the circuit using Multisim and demonstrate your answer to part (b) is correct.

First we will calculate how the changes in temperature will affect the resistance of the sensor:

Now we must find the voltage for the left hand side of this wheatstone bridge.

V1=3V

And now the right hand side of the bridge, this will vary each time as the resistance of the sensor changes. Firstly we will be doing question a) with the sensor representing 115.1Î©:

The reading on the voltmeter will be the difference between those two calculations

b)

Now we continue the calculations for the second value of resistance for the sensor.

With the sensor representing 132Î©:

The reading on the voltmeter will be the difference between those two calculations

c)

Now we continue the calculations for the third value of resistance for the sensor.

With the sensor representing 132Î©:

The reading on the voltmeter will be the difference between those two calculations

#### 2.27 For the Wheatstone Bridge circuit below, what value of R1 will produce a balanced bridge?

Using your calculated answer build the circuit in Multisim and demonstrate your answer is correct.

2.28 A 1m long resistive wire of uniform cross section is connected to a 6V source as shown.

If a sliding contact is placed 0.35m from one end and connected to an unknown e.m.f. then no current is measured on the ammeter.

A) What it the value of the unknown e.m.f.?

This can be solved using the voltage division rule.

### Cite This Work

To export a reference to this article please select a referencing stye below: