Why is the resistance of the circuits input voltage not taken into account when calculating overall resistance of the circuit?

When calculating the resistance in a circuit, one uses Ohm's Law i.e. V = IR with resistors in the circuit.

But why is the resistance of the inputs not taken into account?

For example, if one has a 9V battery with 2 x resistors of 10 ohms each, we can calculate that I = .45 amps. But why is the resistance across the terminals of the battery e.g. 50 M ohms not taken into account in the calculation?

And if it is ever taken into account, under what circumstances?

For that matter, does one take into account the resistance across other components such as capacitors, inductors, op amps, etc?

• Depending on your goal, this is done, but if you just have the circuit in isolation, how could you even do that? – PlasmaHH Feb 7 '17 at 13:33
• All components have ESR , sometimes called Rs in some or DCR in L or RdsOn in Mosfet's or Rce in BJT switches . Each is inversely related to C of battery or capacitor or Pd in diodes or LED's such that the RC product or the Pd*ESR product is constant for any given chemistry. NiCad and LiPo have one of the lowest, coin cells one of the highest ESR – Tony Stewart Sunnyskyguy EE75 Feb 7 '17 at 14:09

What does matter significantly in your hypothetical case is the internal resistance of the battery (which is not a constant- it varies with the charge state of the battery, with temperature, and with the recent history of discharge due to 'polarization' effects). The other thing that matters is the open-circuit voltage of the battery, which varies with internal temperature and with charge state. So we might (as a first approximation) think of the battery as 9.0V ideal voltage source with (say) 3$\Omega$ in series (FYI, the Circuitlab model defaults to 2 ohms- that resistance will rise as the battery discharges). Either will give a more accurate answer than 9.0V ideal with 50M in parallel (the latter which would have no effect other than to drain the battery). Your 20 ohms in series will mean that the battery terminal voltage will drop to 7.8V and the current will be only 0.39A (using 3 ohms).