Basic circuit understanding. If you can't find a parallel equivalent value of two resistors, there is a whole slew of parameters that you won't be able to find.Please explain me why this is used..
So you are saying about this Parallel Equivalent of 2k 2k is equivalent to 1k of series ? Sorry I am very BeginnerBasic circuit understanding. If you can't find a parallel equivalent value of two resistors, there is a whole slew of parameters that you won't be able to find.
Maths is not my strong point - so I cheat.So you are saying about this Parallel Equivalent of 2k 2k is equivalent to 1k of series ? Sorry I am very Beginner
Yes, that is exactly is what I am saying. One equation that you should commit to memory, is E=I⋅R.So you are saying about this Parallel Equivalent of 2k 2k is equivalent to 1k of series ? Sorry I am very Beginner
That's pretty much what I said.Yes, that is exactly is what I am saying. One equation that you should commit to memory, is E=I⋅R.
Okay, first, let's set the voltage to 2 volts. Now, compute the current in the left 2k resistor. It is 2V=I⋅2KΩ, or I=2V/2kΩ, or, I=1mA. Now compute the current in the right 2k resistor. Again, it is 1mA. Now sum the currents at the bottom node. Here you have 1mA + 1mA, or a 2mA sum. Since the sum of the current going out of a network is the same as the amount going in, we know that the voltage supply is supplying 2mA total. To find the equivalent resistance that would also cause the voltage supply to put out 2mA is also found using the formula above, E=I⋅R. Plugging in our values, 2V=2mA⋅R, or R=2V/2mA, or R=1000Ω.
Form an orderly queue please..........................You can also determine the parallel value by calculating the current through each resistor.
Then you add the currents, using basic algebra, to calculate the equivalent single resistance.
For example, for 1k and 2k resistors in parallel:
Itotal = (V/1k + V/2k) = (2V/2k + V/2k) = 3V/2k = V/(2/3k).
Thus the parallel equivalent is 2/3k, the same a calculated from the formula: Rp = (1k * 2k)/(1k+2k) = 2/3k
It's not a competition... or is it?Form an orderly queue please..........................
Okey guys I am very bad at math how ever I am trying to learn it. The question is simple is parallel circuit resistors draw haf of the amount each and is equivalent and why? Is it because it has low current supply?
Has nothing to do with there being a low current supply. Actually your diagram shows a battery and a battery can supply a huge current, at least for a little while.Okey guys I am very bad at math how ever I am trying to learn it. The question is simple is parallel circuit resistors draw haf of the amount each and is equivalent and why? Is it because it has low current supply?
With series resistors you add up the sum of all the individual resistors.Okay first thanks all for help this dummy.. So series resistor gives the same value but in parallel the resistor draws haf of the current each right?
More a case of; imitation is the sincerest form of flattery.It's not a competition... or is it?
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