Illustrated as such:

enter image description here

Am I drawing more current in the first case than the second case? Would this drain the 9V battery faster?

  • \$\begingroup\$ what are the values of the resistors? \$\endgroup\$ – Mark Aug 29 '11 at 4:51
  • \$\begingroup\$ oh, all resistors are 10K \$\endgroup\$ – Shubham Aug 29 '11 at 5:14

The question cannot be answered with the information given.

What matters is the total effective resistance from supply to ground. The resistance can be in one body or in many bodies and still be the same. But, it's obviously possible to have 4 bodies (as in your example) that are not the equivalent of a single resistor.

Current = Volts divided by Resistance.

if the total current flow in your diagrams is the same in each case then the two sets of resistance are equivalent. For reasons which can be explained but which I won't go into here

  • Reffective = 1 / [ 1/R1 + 1/R2 + 1/R3 + 1/R4 ]

So if R5 = Reffective then the situations are identical.
If R5 > Reffective then the four resistor version will drain the battery faster.
If R5 < Reffective then the single resistor version will drain the battery faster.

If R5 = say 1000 ohms then, if R1=R2=R3=R4 = 4000 ohms each then Reffective = 1000 ohms and the situtaions would be identical.


Reffective = 1 / [ 1/R1 + 1/R2 + 1/R3 + 1/R4 ] Reffective = 1/ [ 1/4000 + 1/4000 + 1/4000 + 1/4000]
= 1/[ 4/4000] = 1/1/1000 = 1000 ohms = R5.

Water example:

Water examples are often useful. They are ever perfect but are often very helpful.

Imagine a river with a certain flow rate through a smooth flat bottomed vertical sided channel. Imagine the channel was 16 feet wide and the flow 3 feet deep. Flow rate per second matters but will be the same here in all cases.

If you divided the channel into 2 channels each 8 feet wide with the same material and sides etc you would get an identical result*. [[* In the inconvenient real world the drag from the sides of the channels and other factors may have some effect. These can be ignored for now]].

Now divide the channel into 4 channels each 4 feet wide but otherwise identical as with one channel. This is the equivalent of the resistors being equivalent in your original query.


  • \$\begingroup\$ Sorry forgot to mention theyre all 10K. Ohhh gotcha, the effective resistance of the 4 resistors in parallel would end up being less than the single 10K in the other one...so more current would be drawn. Thanks! \$\endgroup\$ – Shubham Aug 29 '11 at 5:19
  • \$\begingroup\$ Yes - 4 x as much if all the same value . \$\endgroup\$ – Russell McMahon Aug 29 '11 at 5:37

\$ Ohm's Law: U = R \times I \$.

Print it big, frame it and hang it on your den's wall. It's the basis of lots of circuit calculations.
Here we're interested in the current, so we rearrange the equation:

\$ I = \dfrac{U}{R} \$

That means that a higher voltage across the resistor results in a higher current, while a higher resistor value will give a lower current. All resistances are \$10k\Omega\$ and the voltage across each resistor is also the same \$9V\$, so for each resistor

\$ I_R = \dfrac{9V}{10k\Omega} = 900\mu A\$

and the total current in your left schematic is \$4 \times 900\mu A = 3.6mA\$, or 4 times the current of your right schematic.

This higher current will drain your battery four times as fast.

For extra credit:
Any number of parallel resistors can be replaced by a single one. We know the voltage across it, that's \$9V\$, and the total current, that's \$3.6mA\$, then, again according to Ohm's Law

\$ R_{EQUIVALENT} = \dfrac{V}{I} = \dfrac{9V}{3.6mA} = 2500\Omega\$

This means that 4 identical resistors in parallel can be replaced by a single one \$1/4\$ their value, and more general \$N\$ identical resistors in parallel can be replaced by a single resistor \$1/N\$ their value.


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