# Electric

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#1

Part 6b, I can't understand how to do it. Like what should I be applying to this question?
0
6 years ago
#2
(Original post by zed963)

Part 6b, I can't understand how to do it. Like what should I be applying to this question?
a) Remember Kirchoffs current law for the currents in each of the resistor branches. (Current entering node = sum of currents leaving the node)

b) Apply ohms law to find the current in each branch I = V/R (where V = supply voltage because both branches are across the battery terminals and R = series resistance in each branch).

c) Using the current values of b), apply ohms law again (V - IR) to find the pd's across the first two resistors in the question. (A-C 20k ohms and D-F 5k ohms thermistor)

d) Since E-C is the same resistance as A-C and in the same series current path, the pd across both will be the same.

The pd between C-D is then finally calculated from the difference in the pd's across the resistors between E-C (20 kohms) and D-F (5 kohms).
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#3
(Original post by uberteknik)
a) Remember Kirchoffs current law for the currents in each of the resistor branches. (Current entering node = sum of currents leaving the node)

b) Apply ohms law to find the current in each branch I = V/R (where V = supply voltage because both branches are across the battery terminals and R = series resistance in each branch).

c) Using the current values of b), apply ohms law again (V - IR) to find the pd's across the first two resistors in the question. (A-C 20k ohms and D-F 5k ohms thermistor)

d) Since E-C is the same resistance as A-C and in the same series current path, the pd across both will be the same.

The pd between C-D is then finally calculated from the difference in the pd's across the resistors between E-C (20 kohms) and D-F (5 kohms).
Current on the first branch is 0.003 A. So V=IR so 0.003*20000=6v

Current on the second branch is 0.008 A. So V=IR so 0.008*5000=4v

For C-D what would you need to do?
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#4
(Original post by zed963)
Current on the first branch is 0.003 A. So V=IR so 0.003*20000=6v

Current on the second branch is 0.008 A. So V=IR so 0.008*5000=4v

For C-D what would you need to do?
Total current in the circuit is (1/40)+(1/15) = 11/120

1/11/120 = 10.909 ohms

12/10.909 = 1.1 mA
0
6 years ago
#5
(Original post by zed963)
Current on the first branch is 0.003 A. So V=IR so 0.003*20000=6v

Current on the second branch is 0.008 A. So V=IR so 0.008*5000=4v

For C-D what would you need to do?
Both correct.

Using the -ve battery terminal as a reference, what are the voltage potentials above that reference for point C and then point D in the circuit? HINT: points E and F are both connected to the -ve terminal of the battery.

You then have all the information you need to calculate the potential difference between points C and D.

HINT: it's a subtraction.
0
6 years ago
#6
(Original post by zed963)
Current on the first branch is 0.003 A. So V=IR so 0.003*20000=6v

Current on the second branch is 0.008 A. So V=IR so 0.008*5000=4v

For C-D what would you need to do?
Vac+Vce=12V
Vbd+Vdf=12V

since you've calculated the pd across one of the series resistors in each branch you can calculate the pd across the other because they must add up to 12V.

if you know the Vdf and the Vce which are both relative to the potential of the bottom conductor you can work out the Vce

(alternatively you could work from the top supply rail - the important thing is that you get the pd's at C and D relative to the same point in the circuit and don't try to directly compare a pd relative to the top rail with a pd relative to the bottom rail)
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#7
(Original post by Joinedup)
Vac+Vce=12V
Vbd+Vdf=12V

since you've calculated the pd across one of the series resistors in each branch you can calculate the pd across the other because they must add up to 12V.

if you know the Vdf and the Vce which are both relative to the potential of the bottom conductor you can work out the Vce

(alternatively you could work from the top supply rail - the important thing is that you get the pd's at C and D relative to the same point in the circuit and don't try to directly compare a pd relative to the top rail with a pd relative to the bottom rail)

Vce = 6v
Vbd = 8v
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#8
(Original post by uberteknik)
Both correct.

Using the -ve battery terminal as a reference, what are the voltage potentials above that reference for point C and then point D in the circuit? HINT: points E and F are both connected to the -ve terminal of the battery.

You then have all the information you need to calculate the potential difference between points C and D.

HINT: it's a subtraction.
Both branches get 12 volts each.

so is it 8-6=2 ?
0
6 years ago
#9
(Original post by zed963)
Vce = 6v
correct.

(Original post by zed963)
Vbd = 8v
The potentials across the points as you have stated are correct.

However, as Joinedup said in a previous post, you must be very careful to ensure both pd's are referenced from the same point. Your's, as they stand, are not referenced to the same point.

[A quick note about nomenclature: VCE means the pd between the points C and E. The first point 'C' in this example is denoted by the first subscript VC (where you reference the pd). The second point 'E' is denoted by the second subscript VCE (i.e. the potential at that second point referenced to the first point. This is important because it ultimately determines the sign of the voltage in conjunction with the direction of conventional current flow.]

So:

VEC= IECREC = [12/(40x103)] x 20x103 = 6V

Since E and F are at the same potential and both at the lowest potential (zero) in the circuit because they are connected to the battery -ve terminal, then all three act as the same reference point:

VFD = IFDRFD = [12 /(15x103)] x 5x103 = 4V (notice the difference?)

Then finally, the qustion asks for the d between points C - D. i.e. the pd VCD

VCD = VEC - VFD = 6 - 4 = 2V

Which co-incidentally is the same answer you calculated. However, the pd's in a different question could just as easily have been arranged to give an incorrect answer.
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#10
(Original post by uberteknik)
correct.

The potentials across the points as you have stated are correct.

However, as Joinedup said in a previous post, you must be very careful to ensure both pd's are referenced from the same point. Your's, as they stand, are not referenced to the same point.

[A quick note about nomenclature: VCE means the pd between the points C and E. The first point 'C' in this example is denoted by the first subscript VC (where you reference the pd). The second point 'E' is denoted by the second subscript VCE (i.e. the potential at that second point referenced to the first point. This is important because it ultimately determines the sign of the voltage in conjunction with the direction of conventional current flow.]

So:

VEC= IECREC = [12/(40x103)] x 20x103 = 6V

Since E and F are at the same potential and both at the lowest potential (zero) in the circuit because they are connected to the battery -ve terminal, then all three act as the same reference point:

VFD = IFDRFD = [12 /(15x103)] x 5x103 = 4V (notice the difference?)

Then finally, the qustion asks for the d between points C - D. i.e. the pd VCD

VCD = VEC - VFD = 6 - 4 = 2V

Which co-incidentally is the same answer you calculated. However, the pd's in a different question could just as easily have been arranged to give an incorrect answer.
So it's like saying we have a component that takes 6v so the other component in series also has 6v. In the second branch we have 4v and 8v on the other component in series. But before the voltage reaches the second components, what is the difference between the two branches?
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6 years ago
#11
(Original post by zed963)
So it's like saying we have a component that takes 6v so the other component in series also has 6v. In the second branch we have 4v and 8v on the other component in series. But before the voltage reaches the second components, what is the difference between the two branches?
Yes, The word 'potential is the clue here:

Voltage is a measure of potential energy because it's defined as Joules per Coulombs worth of charge.

So 20V means 20 Joules of potential energy is available for every Coulomb of charge that passes any point in the circuit.

When the current meets a resistance, work is done trying to push past - in a way it's a bit like friction.

When work is done as the charge moves through the resistance, it must lose some of it's energy. i.e. some of the potentia energy carried by the charge is converted to heat.

So the potential energy difference (potential difference get it?) is a measure of how much potential energy is lost in that resistance.

That means a 'voltage drop' measured between two points in a current path, is a measure of how much original potential energy the charge had available is converted to other forms of energy.

A pd of 15V between two points in the same conduction path, means the carge flow has lost 15 Joules worth of energy for every coulombs worth of charge that passes between the two points.

In this question it's slightly different:

The current is flowing in two different paths. So all you are doing by stating the potential difference is saying that the first point 'C' still has 6V worth of potential energy (6 Joules per Coulomb) referenced to the battery -ve terminal and, the second point 'D' still has 4V worth of potential energy (4 Joules per Coulomb) again rferenced to the same -ve batter terminal.

The difference in potential between the two points must therefore be 6V - 4V = 2V.
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