show that the equivalent resistance in series is always greater than in parallel combination
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Here’s some information that may be of some more practical use to you. When we put resistors together like this, in series and parallel, we change the way current flows through them. For example, if we have a 10V supply across a 10kΩ resistor, Ohm’s law says we’ve got 1mA of current flowing.

If we then put another 10kΩ resistor in series with the first and leave the supply unchanged, we’ve cut the current in half because the resistance is doubled.

In other words, there’s still only one path for current to take and we just made it even harder for current to flow. How much harder? 10kΩ + 10kΩ = 20kΩ. And, that’s how we calculate resistors in series – just add their values.
To put this equation more generally: the total resistance of N – some arbitrary number of – resistors is their total sum.


Calculating Equivalent Resistances in Parallel Circuits
What about parallel resistors? That’s a bit more complicated, but not by much. Consider the last example where we started with a 10V supply and a 10kΩ resistor, but this time we add another 10kΩ in parallel instead of series. Now there are two paths for current to take. Since the supply voltage didn’t change, Ohm’s Law says the first resistor is still going to draw 1mA. But, so is the second resistor, and we now have a total of 2mA coming from the supply, doubling the original 1mA. This implies that we’ve cut the total resistance in half.

While we can say that 10kΩ || 10kΩ = 5kΩ (“||” roughly translates to “in parallel with”), we’re not always going to have 2 identical resistors. What then?
The equation for adding an arbitrary number of resistors in parallel is:

If reciprocals aren’t your thing, we can also use a method called “product over sum” when we have two resistors in parallel:

However, this method is only good for two resistors in one calculation. We can combine more than 2 resistors with this method by taking the result of R1 || R2 and calculating that value in parallel with a third resistor (again as product over sum), but the reciprocal method may be less work.

If we then put another 10kΩ resistor in series with the first and leave the supply unchanged, we’ve cut the current in half because the resistance is doubled.

In other words, there’s still only one path for current to take and we just made it even harder for current to flow. How much harder? 10kΩ + 10kΩ = 20kΩ. And, that’s how we calculate resistors in series – just add their values.
To put this equation more generally: the total resistance of N – some arbitrary number of – resistors is their total sum.


Calculating Equivalent Resistances in Parallel Circuits
What about parallel resistors? That’s a bit more complicated, but not by much. Consider the last example where we started with a 10V supply and a 10kΩ resistor, but this time we add another 10kΩ in parallel instead of series. Now there are two paths for current to take. Since the supply voltage didn’t change, Ohm’s Law says the first resistor is still going to draw 1mA. But, so is the second resistor, and we now have a total of 2mA coming from the supply, doubling the original 1mA. This implies that we’ve cut the total resistance in half.

While we can say that 10kΩ || 10kΩ = 5kΩ (“||” roughly translates to “in parallel with”), we’re not always going to have 2 identical resistors. What then?
The equation for adding an arbitrary number of resistors in parallel is:

If reciprocals aren’t your thing, we can also use a method called “product over sum” when we have two resistors in parallel:

However, this method is only good for two resistors in one calculation. We can combine more than 2 resistors with this method by taking the result of R1 || R2 and calculating that value in parallel with a third resistor (again as product over sum), but the reciprocal method may be less work.
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Equivalent resistance of a circuitconnected in series is the sum of individual resistors. So when you compare, the equivalent resistance is much greater than the individualresistors. But in case of parallel connection of resistors the equivalent resistance is smaller than the individual resisters.
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