simulate this circuit – Schematic created using CircuitLab

I have a circuit of just a 240 ohm resistors and a 12 v battery. In the circuit the current after the resistors is 0.05. what is the current without the resistors or before the resistors?

Please give full explanation and working out. Im new to this subject and this website.

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    \$\begingroup\$ Please don't post duplicates. This is a duplicate of Find the Current (A) before a ressitor (PS. No idea why, but I fail to flag it as duplicate) (Am also wondering who upvoted it) \$\endgroup\$ – Huisman Jun 5 '19 at 9:17
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    \$\begingroup\$ You have a typo in the schematic at the battery voltage value. \$\endgroup\$ – stowoda Jun 5 '19 at 9:21
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    \$\begingroup\$ Impatience does not pay off at all :/ \$\endgroup\$ – stowoda Jun 5 '19 at 9:22
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    \$\begingroup\$ Its for my assessment and i have a time limit so... please give me an answer \$\endgroup\$ – NEED HELP Jun 5 '19 at 9:26
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    \$\begingroup\$ This site is not devoted to solving assignments and doing homeworks, but it is devoted to help those who want to learn a subject in the realm of electronics and electricity or have complex design issues in the same realm and need a good advice. However, I has already told you, my advice is to use Kirchoff Currents Law. \$\endgroup\$ – Daniele Tampieri Jun 5 '19 at 9:33

I was wondering about answering this question, as you seem to be demanding others to do your work rather than putting in some effort to show some working. But I felt that as you are clearly very early stages of learning about this, I shall give you the benefit of the doubt.

You may be new to this subject, but you should be aware of Ohm's Law: V=IR.

You have said that voltage (V) is 12. You have said that resistance (R) is 240. And you wish to know current (I). Put the numbers into Ohm's law, and you are there.

Current does not disappear. It leaves one terminal of the source, and will flow to the other terminal. As you only have one possible path for the current, if you know the current at point point of this loop, you know it all points of the loop.

If the resistor was to be removed, there are two possible set ups: either the resistor is replaced with a wire, and so you have a short circuit. V=IR still, but R=0 when V=12, so I=infinity. Second option is the resistor is removed completely, and you are left with an open circuit. Again, V=IR, but R is infinity, V=12, so I is 0.

  • \$\begingroup\$ Thank you very much. i understand the concept now. \$\endgroup\$ – NEED HELP Jun 5 '19 at 11:16
  • \$\begingroup\$ @NEEDHELP if it has answered your question, you can accept the answer by clicking the tick between the voting arrows. \$\endgroup\$ – Puffafish Jun 5 '19 at 11:47

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