# Hydroelectricity Analogy Question + finding charge and efficiency help 🌊⚡️

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Alexandramartis

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Hello, I have a question concerning electricity in a form that is analogous to a hydroelectric plant. Although I have answered all of the sections I wondered if someone could offer advice to improve upon my answers or correct any mistakes. Confessedly. I believe that my answer to 2 is a little long and perhaps convoluted. 😁👍

Pumped-storage hydroelectricity allows energy to be stored to prepare for periods of high demand.

Let a high-level reservoir be 500 m above a low-level reservoir.

1) i. What is the useful power output if the water from the high-level reservoir flows at a rate of 100 kg per minute and the turbine is 85% efficient?

Efficiency = useful power output/total power input

Substituting the given values;

85%=useful power output/ 100

useful power output=85*100

useful power output=8500 kg per minute

ii. The generator produces an electromotive force of 120 V; how many coulombs of charge pass every minute?

Q=W/V

W=F*d

W=m*g*h

W=100*9.81*500=490,500 J

Q=490,500/120

Q=4087.5 C

2. Flowing water is often used as an analogy for current. Explain how current, voltage and resistance can be compared with the hydroelectric power example.

In this analogy, charge is represented by the water amount, voltage is represented by the water pressure, resistance is represented by the width between the reservoirs for the water to flow and current is represented by the water flow. Consider two similar hydroelectric plants, each with a high and low level reservoir but one with a wider channel to permit water to flow through.

The pressure of the water pumped from the high to low level reservoirs can represent voltage. The total water in the reservoirs represents charge. The more water in the reservoir, the higher the charge, the more pressure is measured.

Thinking of the reservoir as a battery, as a place to store a certain amount of energy and then release it. If the reservoir is drained a certain amount, the pressure created decreases. This is synonymous to decreasing voltage, like when a torch gets dimmer as the batteries run out. There is also a decrease in the amount of water that will flow through the reservoirs. Less pressure means less water is flowing.

The amount of water flowing through the reservoirs can be thought of as current. The higher the pressure, the higher the flow, and vice-versa. With water, one would measure the volume of the water flowing through the reservoirs over a certain period of time. Whereas in a circuit, one would measure the amount of charge flowing through the circuit over a period of time, current is measured in amperes (defined as 1 coulomb of charge per second).

The volume of water that passes through a narrow width is less than a wider one at the same pressure. This is resistance. The narrow width between the reservoirs "resists" the flow of water through it even though the water is at the same pressure as between reservoirs with a wider channel.

In electrical terms, this is represented by two circuits with equal voltages and different resistances. The circuit with the higher resistance will allow less charge to flow, meaning the circuit with higher resistance has less current flowing through it.

In electrical terms, this is represented by two circuits with equal voltages and different resistances. The circuit with the higher resistance will allow less charge to flow, meaning the circuit with higher resistance has less current flowing through it.

By measuring the same amount of pressure at the end of the reservoirs, when the water begins to flow, the flow rate of the water in the reservoirs with the narrower channel between them will be less than the flow rate of the Thus, the current through the narrower channel is less than the current through the wider channel. If the flow is to be the same through both channel, one would have to increase the amount of water (charge) in the reservoirs with the narrower channel.

This increases the pressure (voltage) at the end of the narrower channel, pushing more water through from the high to low level reservoir. This is analogous to an increase in voltage that causes an increase in current.

The relationship between voltage and current here becomes visible.

Pumped-storage hydroelectricity allows energy to be stored to prepare for periods of high demand.

Let a high-level reservoir be 500 m above a low-level reservoir.

1) i. What is the useful power output if the water from the high-level reservoir flows at a rate of 100 kg per minute and the turbine is 85% efficient?

Efficiency = useful power output/total power input

Substituting the given values;

85%=useful power output/ 100

useful power output=85*100

useful power output=8500 kg per minute

ii. The generator produces an electromotive force of 120 V; how many coulombs of charge pass every minute?

Q=W/V

W=F*d

W=m*g*h

W=100*9.81*500=490,500 J

Q=490,500/120

Q=4087.5 C

2. Flowing water is often used as an analogy for current. Explain how current, voltage and resistance can be compared with the hydroelectric power example.

In this analogy, charge is represented by the water amount, voltage is represented by the water pressure, resistance is represented by the width between the reservoirs for the water to flow and current is represented by the water flow. Consider two similar hydroelectric plants, each with a high and low level reservoir but one with a wider channel to permit water to flow through.

The pressure of the water pumped from the high to low level reservoirs can represent voltage. The total water in the reservoirs represents charge. The more water in the reservoir, the higher the charge, the more pressure is measured.

Thinking of the reservoir as a battery, as a place to store a certain amount of energy and then release it. If the reservoir is drained a certain amount, the pressure created decreases. This is synonymous to decreasing voltage, like when a torch gets dimmer as the batteries run out. There is also a decrease in the amount of water that will flow through the reservoirs. Less pressure means less water is flowing.

The amount of water flowing through the reservoirs can be thought of as current. The higher the pressure, the higher the flow, and vice-versa. With water, one would measure the volume of the water flowing through the reservoirs over a certain period of time. Whereas in a circuit, one would measure the amount of charge flowing through the circuit over a period of time, current is measured in amperes (defined as 1 coulomb of charge per second).

The volume of water that passes through a narrow width is less than a wider one at the same pressure. This is resistance. The narrow width between the reservoirs "resists" the flow of water through it even though the water is at the same pressure as between reservoirs with a wider channel.

In electrical terms, this is represented by two circuits with equal voltages and different resistances. The circuit with the higher resistance will allow less charge to flow, meaning the circuit with higher resistance has less current flowing through it.

In electrical terms, this is represented by two circuits with equal voltages and different resistances. The circuit with the higher resistance will allow less charge to flow, meaning the circuit with higher resistance has less current flowing through it.

By measuring the same amount of pressure at the end of the reservoirs, when the water begins to flow, the flow rate of the water in the reservoirs with the narrower channel between them will be less than the flow rate of the Thus, the current through the narrower channel is less than the current through the wider channel. If the flow is to be the same through both channel, one would have to increase the amount of water (charge) in the reservoirs with the narrower channel.

This increases the pressure (voltage) at the end of the narrower channel, pushing more water through from the high to low level reservoir. This is analogous to an increase in voltage that causes an increase in current.

The relationship between voltage and current here becomes visible.

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i/ will be expecting a value with the unit of Watt - kg is not a unit of power

firstly you'll want to convert the flow rate expressed in mass/min into mass/sec

mgh on the mass per sec you just calculated will give you the input rate of energy conversion (in Joules) per second... which is equal by definition to power in Watts

then apply the efficiency calculation to calculate output.

---

IMO aim for one sentence per mark on long answer questions. when you get to uni you'll be given world count limits and be penalised for exceeding them, so it's worth practising expressing yourself in a few words as necessary.

firstly you'll want to convert the flow rate expressed in mass/min into mass/sec

mgh on the mass per sec you just calculated will give you the input rate of energy conversion (in Joules) per second... which is equal by definition to power in Watts

then apply the efficiency calculation to calculate output.

---

IMO aim for one sentence per mark on long answer questions. when you get to uni you'll be given world count limits and be penalised for exceeding them, so it's worth practising expressing yourself in a few words as necessary.

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Alexandramartis

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#3

(Original post by

i/ will be expecting a value with the unit of Watt - kg is not a unit of power

firstly you'll want to convert the flow rate expressed in mass/min into mass/sec

mgh on the mass per sec you just calculated will give you the input rate of energy conversion (in Joules) per second... which is equal by definition to power in Watts

then apply the efficiency calculation to calculate output.

---

IMO aim for one sentence per mark on long answer questions. when you get to uni you'll be given world count limits and be penalised for exceeding them, so it's worth practising expressing yourself in a few words as necessary.

**Joinedup**)i/ will be expecting a value with the unit of Watt - kg is not a unit of power

firstly you'll want to convert the flow rate expressed in mass/min into mass/sec

mgh on the mass per sec you just calculated will give you the input rate of energy conversion (in Joules) per second... which is equal by definition to power in Watts

then apply the efficiency calculation to calculate output.

---

IMO aim for one sentence per mark on long answer questions. when you get to uni you'll be given world count limits and be penalised for exceeding them, so it's worth practising expressing yourself in a few words as necessary.

1. i. 100 kg per minute = 5/3 ~1.667 kg/s

Do I need to amend my answer to 1i. with the value in kg/s;

ie. 85%=useful power output/ 5/3 kg/s

useful power output=85*5/3

useful power output=425/3 or 141.667 kg/s

I do not think that this appears correct though?

ii. The generator produces an electromotive force of 120 V; how many coulombs of charge pass every minute?

Q=W/V

W=F*d

W=m*g*h

W=5/3*9.81*500=8175 J/s = 8175 W

Q=8175/120

Q= 68.125 C

2. I have tried to shorten my answer (though I know it still requires further brevity), do you think my analogy itself is suitable to compare to a hydroelectric plant?

In this analogy, charge is represented by the water amount, voltage is represented by the water pressure, resistance is represented by the width between the reservoirs for the water to flow and current is represented by the water flow. Consider two similar hydroelectric plants, each with a high and low level reservoir but one with a wider channel to permit water to flow through.

The pressure of the water pumped from the high to low level reservoirs can represent voltage. The total water in the reservoirs represents charge. The more water in the reservoir, the higher the charge, the more pressure is measured.

Thinking of the reservoir as a battery, as a place to store a certain amount of energy and then release it. If the reservoir is drained a certain amount, the pressure created decreases. This is synonymous to decreasing voltage, like when a torch gets dimmer as the batteries run out. There is also a decrease in the amount of water that will flow through the reservoirs. Less pressure means less water is flowing.

The amount of water flowing through the reservoirs can be thought of as current. The higher the pressure, the higher the flow, and vice-versa.

The volume of water that passes through a narrow width is less than a wider one at the same pressure. This is resistance. The narrow width between the reservoirs "resists" the flow of water through it even though the water is at the same pressure as between reservoirs with a wider channel.

In electrical terms, this is represented by two circuits with equal voltages and different resistances. The circuit with the higher resistance will allow less charge to flow, meaning the circuit with higher resistance has less current flowing through it.

I greatly appreciate your help but as you can see I am still a little confused 😳

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Thank you for your reply. Yes I agree and knew that something felt peculiar here but I had just not realised I needed to convert the units to begin with.

1. i. 100 kg per minute = 5/3 ~1.667 kg/s

Do I need to amend my answer to 1i. with the value in kg/s;

ie. 85%=useful power output/ 5/3 kg/s

useful power output=85*5/3

useful power output=425/3 or 141.667 kg/s

I do not think that this appears correct though?

ii. The generator produces an electromotive force of 120 V; how many coulombs of charge pass every minute?

Q=W/V

W=F*d

W=m*g*h

W=5/3*9.81*500=8175 J/s = 8175 W

Q=8175/120

Q= 68.125 C

2. I have tried to shorten my answer (though I know it still requires further brevity), do you think my analogy itself is suitable to compare to a hydroelectric plant?

In this analogy, charge is represented by the water amount, voltage is represented by the water pressure, resistance is represented by the width between the reservoirs for the water to flow and current is represented by the water flow. Consider two similar hydroelectric plants, each with a high and low level reservoir but one with a wider channel to permit water to flow through.

The pressure of the water pumped from the high to low level reservoirs can represent voltage. The total water in the reservoirs represents charge. The more water in the reservoir, the higher the charge, the more pressure is measured.

Thinking of the reservoir as a battery, as a place to store a certain amount of energy and then release it. If the reservoir is drained a certain amount, the pressure created decreases. This is synonymous to decreasing voltage, like when a torch gets dimmer as the batteries run out. There is also a decrease in the amount of water that will flow through the reservoirs. Less pressure means less water is flowing.

The amount of water flowing through the reservoirs can be thought of as current. The higher the pressure, the higher the flow, and vice-versa.

The volume of water that passes through a narrow width is less than a wider one at the same pressure. This is resistance. The narrow width between the reservoirs "resists" the flow of water through it even though the water is at the same pressure as between reservoirs with a wider channel.

In electrical terms, this is represented by two circuits with equal voltages and different resistances. The circuit with the higher resistance will allow less charge to flow, meaning the circuit with higher resistance has less current flowing through it.

I greatly appreciate your help but as you can see I am still a little confused 😳

**Alexandramartis**)Thank you for your reply. Yes I agree and knew that something felt peculiar here but I had just not realised I needed to convert the units to begin with.

1. i. 100 kg per minute = 5/3 ~1.667 kg/s

Do I need to amend my answer to 1i. with the value in kg/s;

ie. 85%=useful power output/ 5/3 kg/s

useful power output=85*5/3

useful power output=425/3 or 141.667 kg/s

I do not think that this appears correct though?

ii. The generator produces an electromotive force of 120 V; how many coulombs of charge pass every minute?

Q=W/V

W=F*d

W=m*g*h

W=5/3*9.81*500=8175 J/s = 8175 W

Q=8175/120

Q= 68.125 C

2. I have tried to shorten my answer (though I know it still requires further brevity), do you think my analogy itself is suitable to compare to a hydroelectric plant?

In this analogy, charge is represented by the water amount, voltage is represented by the water pressure, resistance is represented by the width between the reservoirs for the water to flow and current is represented by the water flow. Consider two similar hydroelectric plants, each with a high and low level reservoir but one with a wider channel to permit water to flow through.

The pressure of the water pumped from the high to low level reservoirs can represent voltage. The total water in the reservoirs represents charge. The more water in the reservoir, the higher the charge, the more pressure is measured.

Thinking of the reservoir as a battery, as a place to store a certain amount of energy and then release it. If the reservoir is drained a certain amount, the pressure created decreases. This is synonymous to decreasing voltage, like when a torch gets dimmer as the batteries run out. There is also a decrease in the amount of water that will flow through the reservoirs. Less pressure means less water is flowing.

The amount of water flowing through the reservoirs can be thought of as current. The higher the pressure, the higher the flow, and vice-versa.

The volume of water that passes through a narrow width is less than a wider one at the same pressure. This is resistance. The narrow width between the reservoirs "resists" the flow of water through it even though the water is at the same pressure as between reservoirs with a wider channel.

In electrical terms, this is represented by two circuits with equal voltages and different resistances. The circuit with the higher resistance will allow less charge to flow, meaning the circuit with higher resistance has less current flowing through it.

I greatly appreciate your help but as you can see I am still a little confused 😳

in order to lift water up to the top reservoir, work is done pushing it up a gravitational potential gradient giving the water gravitational potential energy as height h increases

the amount of the increase of GPE is mgh (as per formula book)

that GPE can be converted to KE by allowing it to flow back down the potential gradient to the lower reservoir

the amount of KE is also given by mgh

the amount of power is the rate at which energy is being converted

so the amount of energy from 1kg of water coming down the pipe is

mgh

1 9.8 500 = 4900 J

and if there was 1kg of water coming down the pipe per second that would be 4900 J/s ... which is an input power of 4900 W to the turbine

so you can feed the mass of water coming down the pipe per second into mgh and you get the input power to the turbine in Watts

---

For your written answer I would write in terms of energy conversion between gravitational potential energy and kinetic energy.

The mass of water is conserved which is analogous to the conservation of charge in the circuit

flow of water is analogous to current

the difference in gravitational potential between the top and bottom reservoir is analogous to electrical potential difference

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Alexandramartis

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#5

(Original post by

The question posed in the OP asks for power and you've still got an answer in mass/time units

in order to lift water up to the top reservoir, work is done pushing it up a gravitational potential gradient giving the water gravitational potential energy as height h increases

the amount of the increase of GPE is mgh (as per formula book)

that GPE can be converted to KE by allowing it to flow back down the potential gradient to the lower reservoir

the amount of KE is also given by mgh

the amount of power is the rate at which energy is being converted

so the amount of energy from 1kg of water coming down the pipe is

mgh

1 9.8 500 = 4900 J

and if there was 1kg of water coming down the pipe per second that would be 4900 J/s ... which is an input power of 4900 W to the turbine

so you can feed the mass of water coming down the pipe per second into mgh and you get the input power to the turbine in Watts

---

For your written answer I would write in terms of energy conversion between gravitational potential energy and kinetic energy.

The mass of water is conserved which is analogous to the conservation of charge in the circuit

flow of water is analogous to current

the difference in gravitational potential between the top and bottom reservoir is analogous to electrical potential difference

**Joinedup**)The question posed in the OP asks for power and you've still got an answer in mass/time units

in order to lift water up to the top reservoir, work is done pushing it up a gravitational potential gradient giving the water gravitational potential energy as height h increases

the amount of the increase of GPE is mgh (as per formula book)

that GPE can be converted to KE by allowing it to flow back down the potential gradient to the lower reservoir

the amount of KE is also given by mgh

the amount of power is the rate at which energy is being converted

so the amount of energy from 1kg of water coming down the pipe is

mgh

1 9.8 500 = 4900 J

and if there was 1kg of water coming down the pipe per second that would be 4900 J/s ... which is an input power of 4900 W to the turbine

so you can feed the mass of water coming down the pipe per second into mgh and you get the input power to the turbine in Watts

---

For your written answer I would write in terms of energy conversion between gravitational potential energy and kinetic energy.

The mass of water is conserved which is analogous to the conservation of charge in the circuit

flow of water is analogous to current

the difference in gravitational potential between the top and bottom reservoir is analogous to electrical potential difference

ie. 85%=useful power output/ total power input

Amount of energy from 1kg of water coming down the pipe = mgh=1 *9.81* 500 = 4905 J

useful power output=85*4905

useful power output=416,925 J ?

ii. The generator produces an electromotive force of 120 V; how many coulombs of charge pass every minute?

Q=W/V

W=F*d

W=m*g*h

W= 4905 J/s

Q=4905/120

Q= 40.875 C

2. I have tried to reword my analogy in the form you have suggested but really have just stated how a hydroelectric power station operates. I do not know how to tie current, voltage and resistance to a comparable nature to gpe and ke? Also you suggested equal representations of charge, current and voltage but not resistance. Would resistance be the width of the passage for the water to pass between the high and low level reservoirs or another variable?

In this analogy the mass of water is analogous to the conservation of charge in the circuit, since it is conserved, the flow of water is analogous to current and the difference in gravitational potential between the top and bottom reservoir is analogous to electrical potential difference.

In hydroelectric power stations, the water in the high-level reservoir possesses stores gravitational potential energy since it is at a higher level than the water on the other side of the dam. As the water falls to the low-level reservoir, this potential energy is converted into kinetic energy, the moving water drives electrical generators, built inside the dam, to produce electrical energy.

I apologise that I am struggling to understand this problem although I can assure you that I am trying to wrap my head around it 👍

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(Original post by

Thank you for your reply. I have tried to amend my answers as you have shown.

ie. 85%=useful power output/ total power input

Amount of energy from 1kg of water coming down the pipe = mgh=1 *9.81* 500 = 4905 J

useful power output=85*4905

useful power output=416,925 J ?

ii. The generator produces an electromotive force of 120 V; how many coulombs of charge pass every minute?

Q=W/V

W=F*d

W=m*g*h

W= 4905 J/s

Q=4905/120

Q= 40.875 C

2. I have tried to reword my analogy in the form you have suggested but really have just stated how a hydroelectric power station operates. I do not know how to tie current, voltage and resistance to a comparable nature to gpe and ke? Also you suggested equal representations of charge, current and voltage but not resistance. Would resistance be the width of the passage for the water to pass between the high and low level reservoirs or another variable?

In this analogy the mass of water is analogous to the conservation of charge in the circuit, since it is conserved, the flow of water is analogous to current and the difference in gravitational potential between the top and bottom reservoir is analogous to electrical potential difference.

In hydroelectric power stations, the water in the high-level reservoir possesses stores gravitational potential energy since it is at a higher level than the water on the other side of the dam. As the water falls to the low-level reservoir, this potential energy is converted into kinetic energy, the moving water drives electrical generators, built inside the dam, to produce electrical energy.

I apologise that I am struggling to understand this problem although I can assure you that I am trying to wrap my head around it 👍

**Alexandramartis**)Thank you for your reply. I have tried to amend my answers as you have shown.

ie. 85%=useful power output/ total power input

Amount of energy from 1kg of water coming down the pipe = mgh=1 *9.81* 500 = 4905 J

useful power output=85*4905

useful power output=416,925 J ?

ii. The generator produces an electromotive force of 120 V; how many coulombs of charge pass every minute?

Q=W/V

W=F*d

W=m*g*h

W= 4905 J/s

Q=4905/120

Q= 40.875 C

2. I have tried to reword my analogy in the form you have suggested but really have just stated how a hydroelectric power station operates. I do not know how to tie current, voltage and resistance to a comparable nature to gpe and ke? Also you suggested equal representations of charge, current and voltage but not resistance. Would resistance be the width of the passage for the water to pass between the high and low level reservoirs or another variable?

In this analogy the mass of water is analogous to the conservation of charge in the circuit, since it is conserved, the flow of water is analogous to current and the difference in gravitational potential between the top and bottom reservoir is analogous to electrical potential difference.

In hydroelectric power stations, the water in the high-level reservoir possesses stores gravitational potential energy since it is at a higher level than the water on the other side of the dam. As the water falls to the low-level reservoir, this potential energy is converted into kinetic energy, the moving water drives electrical generators, built inside the dam, to produce electrical energy.

I apologise that I am struggling to understand this problem although I can assure you that I am trying to wrap my head around it 👍

flow = 100 kg/min = 1.67 kg/sec

height = 500m

efficiency = 85%

power input = energy conversion per second

every second 1.67 kg loses 500m of height in a gravitational field strength of 9.8 N/kg

this might seem like a bit of a mental jump cos it's a continuous flowing liquid rather than separate 1kg masses - but it's actually OK, maybe you could think of 1.67 kg bottles of water going into the pipe at 1 bottle per second and coming out of the bottom at 1 bottle per second (still having the same mass)

mgh = 1.67 9.8 500 = 8180 Joules and since that was a calculation for the mass of water going into the pipe every second it's the number of Joules/second... or number of Watts input to the turbine

efficiency of the turbine = 85%

8180 0.85 = 6960 W power output

electrical power P=VI

I=P/V

=6960/120 = 58A

1 Amp is by definition 1 Coulomb per second so

58C/second... multiply by 60 to get

3480 Coulombs per minute

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#7

(Original post by

Here's what I did

flow = 100 kg/min = 1.67 kg/sec

height = 500m

efficiency = 85%

power input = energy conversion per second

every second 1.67 kg loses 500m of height in a gravitational field strength of 9.8 N/kg

this might seem like a bit of a mental jump cos it's a continuous flowing liquid rather than separate 1kg masses - but it's actually OK, maybe you could think of 1.67 kg bottles of water going into the pipe at 1 bottle per second and coming out of the bottom at 1 bottle per second (still having the same mass)

mgh = 1.67 9.8 500 = 8180 Joules and since that was a calculation for the mass of water going into the pipe every second it's the number of Joules/second... or number of Watts input to the turbine

efficiency of the turbine = 85%

8180 0.85 = 6960 W power output

electrical power P=VI

I=P/V

=6960/120 = 58A

1 Amp is by definition 1 Coulomb per second so

58C/second... multiply by 60 to get

3480 Coulombs per minute

**Joinedup**)Here's what I did

flow = 100 kg/min = 1.67 kg/sec

height = 500m

efficiency = 85%

power input = energy conversion per second

every second 1.67 kg loses 500m of height in a gravitational field strength of 9.8 N/kg

this might seem like a bit of a mental jump cos it's a continuous flowing liquid rather than separate 1kg masses - but it's actually OK, maybe you could think of 1.67 kg bottles of water going into the pipe at 1 bottle per second and coming out of the bottom at 1 bottle per second (still having the same mass)

mgh = 1.67 9.8 500 = 8180 Joules and since that was a calculation for the mass of water going into the pipe every second it's the number of Joules/second... or number of Watts input to the turbine

efficiency of the turbine = 85%

8180 0.85 = 6960 W power output

electrical power P=VI

I=P/V

=6960/120 = 58A

1 Amp is by definition 1 Coulomb per second so

58C/second... multiply by 60 to get

3480 Coulombs per minute

I think I understand everyhting now except I am still struggling with the analogy, could you elaborate on this a litle further. Sorry to keep asking 👍

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Probably saying resistance to flow of water in the pipes is analagous to electrical resistance, in both cases energy is dissipated into the environment would do it...

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Probably saying resistance to flow of water in the pipes is analogous to electrical resistance, in both cases energy is dissipated into the environment would do it...

**Joinedup**)Probably saying resistance to flow of water in the pipes is analogous to electrical resistance, in both cases energy is dissipated into the environment would do it...

2. When an object falls, its gravitational potential energy is changed to kinetic energy;

GPE=KE

mgh=1/2mv^2

here, mgh = 1.67 9.8 500 = 8180 J

The mass of water in the reservoirs is conserved, which is analogous to the conservation of charge in a circuit. The more water in the reservoir, the greater the charge. To generate electricity, water from the high level reservoir passes down through pipes to the low level reservoir, through a generator. As the water travels lower down, it loses gravitational potential energy which is converted to kinetic energy. The amount of water flowing through the reservoirs can be thought of as current. This change in GPE can be thought of as synonymous to voltage and resistance to the flow of water in the pipes is analogous to electrical resistance, as energy is dissipated into the environment.

How can I improve this analogy, could you elaborate why these are suitable analogies:

- electrical resistance=resistance to flow of water in the pipes

-charge=conserved mass of water

-voltage=difference in gravitational potential between the top and bottom reservoir

-current=water flowing

I see that the amount of water in the reservoirs is a fitting analogy for the conservation of charge since this too is conserved. I understand that the flow of water is a fitting description of current, since current is defined as the rate of flow of electric charge. Voltage is defined as the difference in electric potential between two points, so the depicting of p.d. as the difference in GPE between the upper and lower reservoirs is synonymous to this. And electrical resistance opposes the flow of current, which equates to the difficulty in water flowing.

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