simulate this circuit – Schematic created using CircuitLab

  • 2
    \$\begingroup\$ What work have you done so far? \$\endgroup\$ – Null Oct 29 '14 at 14:08

For the 5V source you get: \begin{align} V_{O,5V} = 5V * \frac{2 \Omega \parallel 3\Omega}{4 \Omega + (2 \Omega \parallel 3\Omega)} \approx 1,15V \end{align}

For the 10V source you get: \begin{align} V_{O,10V} = -10V * \frac{4 \Omega \parallel 2 \Omega}{3 \Omega + (4 \Omega \parallel 2 \Omega)} \approx - 3,07V \end{align}

So in total you get: \begin{align} V_O = V_{O,5V} + V_{O,10V} = 1,15V + (-3,07V) = -1,92V \end{align}

Background: For using superposition you have to replace all ideal voltage sources with short circuits and ideal current sources with open circuit (Hope this translation is correct?). Then you calculate the value you want to know. Then you repeat the steps with each source given in the circuit and then add the results.


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