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enter image description here
I have tried to solve this diagram shown above to get total current:
Since $$V_B = V_1 + V_2 + V_3$$ $$12 = V_1 + 4 + V_3$$ $$V_1 + V_3 = 8$$ $$V_1 = 10I$$ $$V_3I = 12$$ $$I = \frac{V_1}{10}$$ $$I = \frac{12}{V_3}$$ $$\frac{V_1}{10} = \frac{12}{V_3}$$ $$V_1V_3 = 120$$ $$V_1 = \frac{120}{V_3}$$ By substituting in $$V_1 + V_3 = 8$$ $$V^2_3 - 8V_3 + 120 = 0$$ I got stuck here and I couldn't complete the solution because I solved this equation with calculator I got imaginative numbers.

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  • \$\begingroup\$ Please use _ character to denote the subscript in MathJax \$\endgroup\$
    – Eugene Sh.
    Jun 14, 2018 at 19:22
  • \$\begingroup\$ $$\begin{align*} V_\text{B} &= I\: \left(R_1+R_2+R_3\right) \\\\ V_\text{B} &= I\: \left(R_1 + \frac{V}{I} + \frac{P_w}{I^2} \right) \\\\ I\: V_\text{B} &= R_1 \: I^2 + V \: I + P_w \\\\ 0 &= R_1 \: I^2 + \left( V-V_\text{B} \right) \: I + P_w \end{align*}$$ The roots are complex conjugates and not real. \$\endgroup\$
    – jonk
    Jun 14, 2018 at 19:42

2 Answers 2

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The situation depicted in the circuit is not possible.

Consider for a minute only the 12v source and the load resistor apparently dissipating 12 watts. If these were the only two components, the current through the resistor would be 1 amp.

Now put the 10 ohm resistor back in the circuit. What is the voltage drop required to push an amp through 10 ohms? Clearly enough that there can no longer be 12 watts dissipated across the other resistor as indicated.

Put the allegedly measured 4v drop back into the circuit and it gets even further from reality.

You have either a trick question, or a miscopied question.

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    \$\begingroup\$ Try instead with a decimal point Pd=1.2W then it can be solved and use Ref. Designations R1,R2,R3 \$\endgroup\$ Jun 14, 2018 at 19:40
  • \$\begingroup\$ @Chris Stratton So it's impossible to determine the current? \$\endgroup\$ Jun 14, 2018 at 19:44
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    \$\begingroup\$ @MohamedMagdy, it's impossible for for all the stated conditions to be true: 12 V source, R1 = 10 V, R2 has 4 V across, and R3 consumes 12 W. \$\endgroup\$
    – The Photon
    Jun 14, 2018 at 20:08
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    \$\begingroup\$ Maybe R3 has a negative resistance and its actually providing 12 W rather than consuming 12 W? \$\endgroup\$
    – The Photon
    Jun 14, 2018 at 20:09
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I believe that your calculator result with complex numbers wasn't all wrong. I tried to solve it on paper and ended up with complex numbers as well. With Kirchhoff's Laws in a DC circuit the given values lead to a contradiction and cannot be solved. In an AC circuit you can have R3 as a capacitor or an inductor. So I ended up with this formula. $$ (I-0.4A)^2 = -1.04A^2 $$ Solve to I and put it in your element equations. (We don't know the frequency of the circuit, though..)

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    \$\begingroup\$ We do know the frequency: it's zero. There's a 12 V battery symbol. You also need to explain how to get 12 W out of a capacitor or inductor. \$\endgroup\$
    – Transistor
    Jun 14, 2018 at 21:36
  • \$\begingroup\$ I agree with you that all the schematic symbols point in the direction of a DC circuit. I only made the statement of the possibility of AC behavior while implying that it would be a trick question then. In DC the circuit cannot be solved. \$\endgroup\$ Jun 14, 2018 at 21:47
  • \$\begingroup\$ Capacitors and inductors aren't usually designated R. \$\endgroup\$ Jun 14, 2018 at 22:45

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