Okay, don't hate me too much or criticize me here. I'm just new to Proteus and I don't know which part I got it wrong but if you'll help me out, I'd really appreciative it!

In the first picture, it's supposed to have a total current of 10 A and then 6.25 A, 1.25 A and 2.5 A respectively but it keeps me giving this weird numbers. I know they're quiet close however my professor won't accept it.


Same problem here too, obviously it's supposed to give out 125 V but why is it giving 124 V? enter image description here

  • 1
    \$\begingroup\$ Hello and Welcome. For starters, could you please organise the schematics? conventions call for positive voltage on top, ground on bottom. \$\endgroup\$
    – Sclrx
    Nov 22, 2017 at 13:50
  • \$\begingroup\$ I do not use Proteus at all but have used many other circuit simulators. First thing I notice is that you do not have a ground symbol. Why is that important? Because it tells the simulator: this is 0 Volt. In your setup it has to make an assumption so it could assume your circuit is at 1.23456 Million Volt then it will calculate the voltage across your resistors by subtracting two large numbers, that can introduce inaccuracy errors which you see. So add a ground and try again. Add a ground to each circuit on the page. \$\endgroup\$ Nov 22, 2017 at 13:53
  • \$\begingroup\$ What are the characteristics of B3 and B4? \$\endgroup\$
    – Finbarr
    Nov 22, 2017 at 13:54
  • \$\begingroup\$ Also I see that you have several independent circuits in your sheet as well. That is OK as long as they're not floating but in your sheet they are floating since you do not have grounds. \$\endgroup\$ Nov 22, 2017 at 13:57
  • \$\begingroup\$ I bet the voltage source has a small series resistance built-in, and its default value is 0.1Ω. \$\endgroup\$
    – W5VO
    Nov 22, 2017 at 13:58

1 Answer 1


While I am not familiar with Proteus, some simulation packages will include a series resistance with the ideal voltage source. This is usually not an issue and helps the simulator solve "hard" SPICE problems like shorting out ideal voltage sources.

Going by the numbers, you expect 12.5 A, which corresponds to a resistance of 12.5 Ω. The actual current is 9.92 A, which means that the ideal voltage source is actually feeding 12.6 Ω. Your second figure confirms this, as the voltage across each resistor is 124 V, which is a reasonable figure if there is a 0.1 Ω resistor in series (1.2499 V rounded to 1 V).


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