Problem 12 exam 1

Practice question for ECE201: "Linear circuit analysis I"

By: Anonymous

Topic: Nodal analysis


The value of voltage $ v_1 $ in the following circuit is:



Assuming that ground is at the bottom of the circuit, as we usually do, we want to find the voltage of the other two nodes A and B. Our unknowns will be $ V_A $ and $ V_B $ and the desired voltage $ v_1 $ can be expressed as $ v_1=V_B-V_A $.

The voltage $ V_A $ is easy to find, since node A is connected to a grounded voltage source. However, you have to be careful with the polarity of this source, since it has the plus side connected to ground and the minus side connected to the node. Then $ V_A=-2V $.

Now we need to write a nodal analysis equation to node B:

$ \begin{align} \frac{V_B-V_A}{2}+0.5v_1+\frac{V_B}{4}-2= & 0\\ \frac{V_B-(-2)}{2}+0.5(V_B-(-2))+\frac{V_B}{4}-2= & 0\\ V_B= & 0 \end{align} $

Finally, $ v_1=0-(-2)=2 $.

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