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I'm trying to self-study this problem but I got stuck. enter image description here

What I tried to do:

The equivalent resistor R1 for the 12 and 4 ohm resistors is 3 (12x4/12+4). The equivalent resistor R2 for the 4 and 12 ohm resistors is 3 (4x12/4+12).

R1 and R2 would be in series now and their equivalent resistor would be 6 Ohms (3+3).

The Is would be 10A (I'm not really sure). Vs would be then 6 x 10 = 60 V. I couldn't figure out why my Is calculation is wrong and why the correct answer is 20A.

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  • \$\begingroup\$ Yes, 100% sure. \$\endgroup\$
    – My Name
    Sep 19, 2014 at 13:18
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    \$\begingroup\$ Is the 10A a constant current source? \$\endgroup\$
    – MrPhooky
    Sep 19, 2014 at 13:26
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    \$\begingroup\$ (d) Is = 20V...HUH??? \$\endgroup\$
    – EM Fields
    Sep 19, 2014 at 14:11
  • \$\begingroup\$ @elliotdawes yes constant. \$\endgroup\$
    – My Name
    Sep 19, 2014 at 14:12
  • \$\begingroup\$ @EM Fields, it's either: they're trying to mislead students or it's a typo. \$\endgroup\$
    – My Name
    Sep 19, 2014 at 14:13

3 Answers 3

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Put some currents on the circuit: -

schematic

simulate this circuit – Schematic created using CircuitLab

Then recognize that where the 10 amps is flowing is at the potential V/2. This leads to: -

10 amps = \$\dfrac{\frac{V}{2}}{4\Omega} - \dfrac{\frac{V}{2}}{12\Omega}\$

10 amps = \$\dfrac{3V - 1V}{24\Omega}\$

10 amps = \$\dfrac{V}{12\Omega}\$ which means V = 120 volts

Can you take it from there given that each resistor has 60 volts across it.

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    \$\begingroup\$ @AmitHasan I don't believe that the wire carrying 10A is meant to be a current source driving 10A. Call it gut feeling!! \$\endgroup\$
    – Andy aka
    Sep 19, 2014 at 15:20
  • \$\begingroup\$ Well, if I put an ammeter in that 10A conducting wire when Vs=120V and Is = 20A then the ammeter shows 0A. Why? Shouldn’t it be 10A? \$\endgroup\$
    – Amit Hasan
    Sep 19, 2014 at 15:26
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    \$\begingroup\$ @AmitHasan whatever the curent that flows from the voltage source, most of it flows thru the 4 ohm resistor and if the top 12 ohm and 4 ohm swap places in the circuit then you'll see no current in the horizontal wire - I bet that's what you tried? \$\endgroup\$
    – Andy aka
    Sep 19, 2014 at 15:29
  • \$\begingroup\$ Yes and Brilliant. I wish I had a teacher like you? \$\endgroup\$
    – Amit Hasan
    Sep 19, 2014 at 15:32
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    \$\begingroup\$ @AmitHasan - I expect an upvote for that amount of respect LOL!! \$\endgroup\$
    – Andy aka
    Sep 19, 2014 at 15:33
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So it's a nice bridge, I made R1= 12 ohm resistor and R2 = 4 ohm. Because of the symmetry the same current will flow through both 4 ohm resistors.
Then doing current conservation.

I2= 10 + I1

and the voltage drop

I1*R1 = I2*R2.

which gives

I2 = 3*I1

plug into first and solve.

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  • \$\begingroup\$ Can you draw how you concluded the current by the symmetry? I didn't understand. \$\endgroup\$
    – My Name
    Sep 19, 2014 at 13:57
  • \$\begingroup\$ Hmm, OK the voltage at the midpoint of the bridge has to be 1/2 of the total. (can you "see" that?) So then top and bottom look exactly the same.. same resistors same voltage drop.. it's got to be the same current. Maybe you can write out all the equations and show this is the case. (Assume different currents see what happens.) \$\endgroup\$ Sep 19, 2014 at 14:06
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There's a trick to explain current flows. Schematic below shows two equivalent circuits, the original is on the left, and modified is on the right where two resistors replaced with pairs connected in parallel.

schematic

simulate this circuit – Schematic created using CircuitLab

There are two wires marked "X". They connect three branches of series resistors, and as these brances are balanced (I tuned resistor values such), there should be no current at "x" wires. So current of 10 A flow through R6 and R9. This gives us voltage between circuit terminals of 10 A × (6+6) Ohm = 120 V.

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