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Before asking my question,I want to talk about certain points 1.Lets suppose there is a battery of 12 volts,this means that the potential difference between the negative and positive terminal of the battery is 12V,and that every 1 coulomb of charge gets 12 joules of energy when it is in the negative terminal and spends it in travelling from the negative terminal to the positive terminal through the wire,and in the resistance as well 2.In Parallel circuit,the voltage across all the resistors is the same as the voltage of battery,let's suppose 12 volts,and the current gets divided at point 2,3,4,5,6,7,8

So now,let's assume there is 1c of charge flowing in the circuit,and it has been given 12 joules of energy from the battery,then,at point 2,it's going to get divided,into let's assume 0.5 C, alright?then,it has 6 joules of energy,but,the potential difference across the resistor is of 12 volts, meaning it is going to lose 12 joules of energy when passing through the resistor,what does this mean?how can a charge going through the resistor lose 12 joules of energy when it only has 6 joules of energy?? please don't focus too much on numbers here,they might be wrong or something,I just used random numbers, please tell the concept,what will happen here?can someone give a detail insight I'm only in 10th grade and 15 years old,so kind you,avoid using concepts which haven't been taught to me

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"so now,let's assume there is 1c of charge flowing"

This is the problem. You can't do this for what you drew. By assuming this, you have turned your voltage source into a current source.

In a voltage source, the voltage is fixed and the circuit determines how much charge/current flows follows from that. In a current source, the current/charge flowing is fixed and the voltage that appears follows that.

It is like saying F=ma in a problem with a block of known mass and then assuming X Newtons and Y acceleration, ignoring that the force and resulting acceleration are related to each other through the mass and therefore cannot be chosen independently of each other. If you do, you end up where the numbers don't work out just as you have.

Using your words, this is the correct way:

12V appears across every resistor. The value of R1, R2, and R3 requires that 1C, 2C and 3C of charge flow through the resistor if that voltage appears across it. That means the battery must be producing 1+2+3=6C charge total. So 6C flows from the battery with 12V*6C = 72J. This is split up so R1, R2, R3 dissipates 12J, 24J, 36J respectively.

If your circuit had a current source that was 1C, then you could mostly say what you said except you cannot say that 12V appears anymore. The voltage that appears would be dependent on the equivalent resistance of R1,R2,R3 put together in parallel.

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  • $\begingroup$ Quoting the OP would make them think that a battery "produces" charge $\endgroup$ Feb 27, 2022 at 21:19
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    $\begingroup$ @Mehmer I am not overly concerned with that at the moment. They do use Joules and not Watts so they do not seem completely ignorant to the difference and I trust they will eventually compensate for that. $\endgroup$
    – DKNguyen
    Feb 27, 2022 at 21:21
  • $\begingroup$ So basically,the maths would disobey what I said,right? $\endgroup$ Feb 27, 2022 at 22:07
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    $\begingroup$ @AakashGarain Yes, because you set up voltage and current values independently when they are not independent. $\endgroup$
    – DKNguyen
    Feb 27, 2022 at 22:12
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it has 6 joules of energy,but,the potential difference across the resistor is of 12 volts, meaning it is going to lose 12 joules of energy when passing through the resistor

No. The potential difference is 12 V, so that means that 1C passing through would lose 12 J. But the 0.5C that passes through would lose 6 J. Voltage is energy per charge.

The other 0.5C would lose 6J in passing through the other resistors.

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  • $\begingroup$ So basically, the 0.5 c of charge would give only 6 joules of energy because it is capable of giving only that much,right?if it had more energy,it would have given more to the resistor, right? $\endgroup$ Feb 28, 2022 at 7:37

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