Blog #37 - Calculating Another Circuit Using Loop Analysis

As part of my problem series in this video, I will be Calculating another circuit using Loop Analysis. Once I am finished with this series of problems I will be posting them in my Stan store, at this web address…https://stan.store/GVB

Find V0 in this circuit using the Principle of Superposition.

To apply superposition, we consider the contribution that each source independently makes to the output voltage V0 . In so doing, we consider each source operating alone and we replace the other source(s). Recall, that in order to replace a voltage source, we replace it with a short circuit since the voltage across a short circuit is zero. In addition, in order to replace a current source, we replace the current source with an open circuit since there is no current in an open circuit.

Consider now the voltage source acting alone. The network used to obtain this contribution to the output V0 is shown above…This shows only the part of V0 that is the contribution due to the 12V source. Using voltage division, that contribution to V0 is… 

V’12 is -12V[4kΩ/(4kΩ+6kΩ+8kΩ)]…(the minus sign is due to the polarity of V12)… and all of that…= -8/3 Volts…The current source contribution to V0 is obtained from this network now shown…Using current division, we find that the contribution from the 6 mA current source, is …6mA[6kΩ/(6kΩ+8kΩ+4kΩ)] = 2 mA which means the voltage contribution of the current source is the voltage drop across the 4kΩ resistor which according to Ohm’s law is = 4kΩ*I’6 mA = 8 Volts…Then superposition states that V0 = V’12 + V’6mA which calculate to…16/3 or 5.33 Volts

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