# Energy efficiency in an electrical circuit

I’m in high school and we are going to do this laboratory in which we will have to make two solutions at different concentrations and determine which has the greatest electrical resistance in a circuit (we will connect it with wires to a 9V battery, calculate its amperage and voltage to determine the resistance of each) However, we must also mention the electric energy, the electric power and the energy efficiency in our analysis (lab report). We have to explain why energy can be released as heat in a circuit using the output/input energy formula. We also have to explain why a more concentrated solution has less resistance (using the notion of ions) while stating why energy can be lost through the circuit (using P=UI and E=PdeltaT).

And so my questions are:

1. how can heat be released in this type of circuit (without a resistor but with a solution)?
2. If we do not take heat into account, how can this affect our results or the energy as such.
3. Finally, could you explain what makes a solution more conducive?
• Hi Nina! As an engineer, I must say this is a pretty solid project, and you're doing a good job at describing it. Welcome to this community! Commented Jun 8, 2021 at 23:07
• Now, what we'd have to ask you (like anyone else) to really help you is where exactly you're stuck – I'm pretty sure you've already got some suspicions on 1. and 3., is my guess! But that's only a guess, and I don't want to start explaining things like waaaaay to low-level for you, if you already know most of it, but I also don't want to skip all the important details that I think you already know, but which you might be missing. So, what's your suspicion? What's in the solution? What did you learn up to now about solutions of that kind? What makes them special according to what you know? Commented Jun 8, 2021 at 23:10
• If you have a DMM meter you can predict the current if using the liquid as a resistive load and if >1k light an LED in series with the correct polarity. Otherwise pure water won’t register on a 20 Megaohm scale. All batteries also have resistance inverse to energy capacity . Then rises sharply e.g. 9V <10 ohms rises towards 1k when dead. Batteries are like dense capacitors precharged but only tolerate 10 to 15% in voltage as heat from losses. All batteries have an initial mAh or Ah rating which is normally rated for a 20 hour load. Commented Jun 9, 2021 at 0:31
• Hi! I’ve got ideas for number one/ three. I know that a solution is more conductive when it is more concentrated. Salts, acids and basic solutions conduct electricity (electrolytes) and the more they are concentrated, the more ions there are and it is easier for electricity to pass through. I also know that heat can be released from a system that is not “closed”. Heat is released (energy can’t be created or can’t just vanish either, it has to be transformed through heat). So what I was really wondering is how does this whole process happen? How does electric energy be transformed into heat?
– Nina
Commented Jun 9, 2021 at 0:58
• "If we do not take heat into account, how can this affect our results or the energy as such." -- Have you been taught about the Law of Conservation of Energy? Commented Jun 9, 2021 at 2:11

1. Why do you think a solution does not have resistance? Do you believe a "resistor" is simply a manufactured electrical device? Can a piece of graphite from a pencil act as a resistor?
2. There are electrochemical effects in which a solution [and electrodes!] can act as a rechargeable cell, storing electricity, or create electrolysis, changing electrical energy to chemical (potential) energy, in addition to changing electrical energy to heat. You'll need to take into account such factors as the electrical potential (voltage), the composition of the electrolyte and the composition of the electrodes.
3. There are many references online describing conductivity of a solution, such as Andy Connelly', or Wikipedia's. What work have you done researching this?
• Hi! So I’ve done research about conductivity (what solutions can conduct electricity), and I know that heat can be released while using a device depending on the efficiency of the device. What I really want to understand is how this formula ( W out / W in * 100) can be applied in my laboratory. In what context can energy efficiency be applied in electrical circuits. (Our circuit is composed of a solution, a battery, wires and we will use a multimeter to calculate the amperage and voltage).
– Nina
Commented Jun 9, 2021 at 1:06
• Hint: "efficiency" means how much useful work is done compared to the amount wasted. This can be expressed as a percentage, q.v. Commented Jun 9, 2021 at 18:03
• Efficiency is about the proportion of 5he energy you USE is doing the thing you want it to do. Usually the rest ends up wasted as heat, noise etc. Eg a motor can be 90% efficient if you can measure (electrically) the power in and also the power on the output shiaft (mechanically!) and find 90% of the value., the other 10% being lost as heat. Efficiency is about what you WANTED the energy to do. A very inefficient motor is probably a very EFFICIENT heater! What is the energy in your setup SUPPOSED to be doing! Measure that (maybe Non-electrical) then compare to the electricity you put in.
– Dan
Commented Jan 29, 2022 at 14:03

Efficiency is going to require a little thought. It's the ratio of useful energy expended to total energy expended. In a Tesla, it's obvious what the useful energy is; it's the forward motion of the car. In your experiment, you'll have to decide, what is the useful outcome.

I can think of a couple of ways :

1. Electrolysis requires some minimum voltage for the reaction. (This is closely related to the voltage a 1-cell battery can generate, both are the chemical energies of the solutions and electrodes). Any voltage above that is wasted (just creates heat). So if you can find the minimum voltage (V) for the reaction to work (conduction starts), efficiency using a 9V battery is V/9 * 100%. (Or whatever you measure from the battery instead of 9. If you can't find an adjustable voltage, maybe you can do it in 1.5V steps with AA cells.)

2. Electrolysis in many solutions (*) releases hydrogen at one electrode and oxygen at the other. If you can catch these bubbles in an upside down test tube filled with water, and note the volume of gas, (should be twice as much H as O) you can, with help, work out how much energy is stored by that volume of hydrogen. (This is going to happen a lot with cheap solar power, and may end up powering aircraft one day soon.) So that's a good measure of useful energy.
(*) Not all solutions : copper sulphate releases copper rather than hydrogen : get the current right and you can copper plate the anode terminal, and polish it later.

3. You might consider any heating in the solution to be useful; or you might consider it to be a waste.

• Thank you very much for your answer. I don’t think we’ll catch these bubbles, but I’ll definitely add that in my lab report as the lost energy (energy that we didn’t take into account).
– Nina
Commented Jun 9, 2021 at 11:23
• If the bubbles are the lost energy, then the heating is the useful energy, right? I was thinking it the other way round, but it's your experiment and so you decide what is useful and what is waste. In industry, you'd want to collect and sell that waste!
– user16324
Commented Jun 9, 2021 at 11:28