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you

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hello in this video we are going to solve simple circuit problems, simple circuits

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are those that only have a single receiver and in this case they are going to be

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electrical resistors so we are going to solve the first problem the statement says given the circuit

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of the figure knowing the value of the vp battery are 10 volts the current intensity and are 2

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amps calculate the value of the resistance r of the circuit well what we are going to do now

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the first thing is to put the law of r is going to be equal to the voltage of the battery divided by the

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current intensity and we are going to square it

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now the next thing is going to be to replace the values ​​we have r no

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so we leave it as it is we have vp if according to the statement they are 10 volts

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we have the current intensity if they are 2 amps with which we are going to put it down here r is going to

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be equal to 10 by 2 5 and as the semidenomial resistors we put their corresponding humidity

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now what we are going to do is square the solution this is always done both in mathematics

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as in technology and next year in physics and chemistry it is very good now we are going to see another problem

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here we already have the second problem in this case or as you see it is the same scheme and the

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statement given the circuit of the figure calculate the value of the intensity and if the resistance r

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is 5 ohms and the vp battery is worth 60 volts because nothing we do the same we are going to put the

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law of today again we put r is going to be equal to the voltage of the vp battery divided by the current

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and we will square it again to distinguish it from what we are going to do next

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we put a comma so that there are no errors and we replace the values ​​we give r if 5 or less because we put it

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we have the value of the vp battery if 60 volts we are going to make the line a little more it gives us

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the current intensity it is not just what it asks us very well because we clear it is there that it is

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dividing it has to pass here multiplying and these 5 miles that are multiplying pass

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dividing because this would be left to us in this way and it will be equal to 60 volts divided by

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5 or less we go down here and it will be equal to 65 they will be 12 amps

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this would be the solution and as I said in the previous problem we are going to square the solution

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and so it would be solved well now we are going to do a third problem more we are going to look for it

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here we have the third problem as you can see we continue to maintain the same scheme we are going to

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see what the statement says given the circuit of the figure calculate the value of the vp battery if

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the resistance r is of isomers and the intensity and is equal to 4 amps that is to say they are not

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asking for the value of the battery because nothing we put the law of or and we start as

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always r is going to be equal to vp divided by the current intensity and as the previous time

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we are going to square it well we substitute we know r if it tells us that they are 6 or mine because we are going to place them

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with or vp it is not just what they ask me I know the current intensity if how much is 4

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4 amps because we have to clear this as we clear it because in this case 4 amps that is

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here dividing happens by multiplying with which I would have that vp is going to be equal to 6 or mine by

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4 amps the product would be 6 by 4 24 and the units in which the voltage is measured would be in

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this case volts the small water then this would be like this vp will be equal to 24 volts and

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like the previous programs we will continue to square it and with this we would already have the three

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types of problems that we can find in the simple circuits formed by a single

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receptor that in this case is an electrical resistance I hope it will serve you