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Consider the below schematic

schematic

simulate this circuit – Schematic created using CircuitLab

How would we find the current flowing across the short circuit? I know what a short circuit is (no voltage) and I know how to find the current across all the resistors using kirchoffs laws. However, applying that to the circuit doesn't seem to add up.

Is there a trick to this?

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  • \$\begingroup\$ Algebraic sum of current will still be the current through series combination of parallel combination of resistors. \$\endgroup\$ – Umar Feb 18 '17 at 3:46
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    \$\begingroup\$ It is same node, you need 2 nodes to find I. \$\endgroup\$ – Electroholic Feb 18 '17 at 4:01
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    \$\begingroup\$ Can you figure out how much current is flowing through R4? And how much is flowing through R2? If those two are different, the difference has to come from current flowing through the short. \$\endgroup\$ – The Photon Feb 18 '17 at 4:01
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A alternative way is to view the circuit being fed from two identical voltage sources each of 12 volts. You have 12 volts feeding R1 and R3 from the left and another 12 volts feeding R2 and R4 from the right. The top wire connecting R1 and R2 can now be broken.

Next convert the 12 volts on the left, R1 and R3 to a simpler voltage source with a single resistor output impedance. This is done by Thevenins/Nortons transformation but the upshot is that to do it you: -

  • Work out what the open circuit voltage is at the junction of R1 and R3
  • Work out what the source impedance would be.

For the first bullet, the open circuit voltage is 12 volts x R3/(R1+R3) = 8 volts. For the 2nd bullet the impedance is R1||R3 = 16.667 ohms.

Repeat for the right hand circuit and you get a voltage source of 6 volts with a series impedance of 37.5 ohms. Left hand and right hand circuits now connect like this: -

schematic

simulate this circuit – Schematic created using CircuitLab

The net voltage across those two resistors is 2 volts and therefore the current is 36.92 mA.

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  • \$\begingroup\$ Thank you so much. Why can't I use Kirchoff current and voltage lows for this? (can't draw a circuit showing the current flow) but we would have 6 currents. The currents I have are as follows: 12V: .274 A R2: .69 A R1: .206 A R3: -.137 A Short: .069A R4: .137 A. However, looks like this isn't correct. I'm I doing this wrong, or is there a reason why this doesn't work? I'm on my first semester circuits class and we haven't learned about Thevenins/Nortons transformation yet. \$\endgroup\$ – Joe Jackson Feb 19 '17 at 2:25
  • \$\begingroup\$ your individual currents appear to be too high at first glance. Maybe show your working in an answer but make it clear that you will delete this answer so as not to attract down votes. Maybe, instead, add the calcs to your original question. \$\endgroup\$ – Andy aka Feb 19 '17 at 9:55
  • \$\begingroup\$ BTW I only see three currents to solve, I1, l2 and Ix, where Ix is the current through the short. I1 and I2 flow through R1 and R2 respectively. \$\endgroup\$ – Andy aka Feb 19 '17 at 10:01

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