Question 2.3: Determine the value of voltage source current in the paralle......
Determine the value of voltage source current in the parallel circuit below.
Similar to Example 2.2, we will present two different approaches for determining the value of unknown source current. The first approach simply utilizes Ohm’s law and the parallel circuit simplification method. The second approach, on the other hand, utilizes KCL and “nodal” analysis technique.
APPROACH I
Reduce or simplify the given circuit to a voltage source and equivalent resistance R_{eq} . Since R_1, R_2, and R_3 are in parallel, application of Eq. 1.13 yields:
R_{eq}=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+…+\frac{1}{R_a}} (1.13)
R_{eq}=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}R_{eq}=\frac{R_1R_2R_3}{R_1R_2+R_2R_3+R_1R_3}
This simplifies the given parallel DC circuit as follows:
Next, Ohm’s law is applied to determine the source current:
I=V_s\left\lgroup\frac{R_1R_2+R_2R_3+R_1R_3}{R_1R_2R_3}\right\rgroupAPPROACH II
This approach is premised on the application of KCL to the given circuit after the node had been identified and circuit has been annotated with voltage designation, voltage polarity, branch currents, and current directions. See the circuit diagram below:
Subscribing to the definition of a node as a point where three or more conductors merge, the shaded segment in the diagram above is designated as the node for this circuit. Next, before KCL can be applied to determine the source current, the indi-vidual currents, through each of the resistors, need to be defined – using Ohm’s law – in terms of the specific resistance values and the voltages around them:
I_1=\frac{V_1}{R_1} I_2=\frac{V_2}{R_2} I_3=\frac{V_3}{R_3}Then, application of KCL at the designated node yields the following equation:
I=I_1+I_2+I_3Substitution of the values of branch currents, as defined earlier, yields:
I=\frac{V_1}{R_1}+\frac{V_2}{R_2}+\frac{V_3}{R_3}At this juncture, it is important to note that when circuit elements are in parallel – as is the case with R_1, R_2 and R_3 – their voltages (or voltage drops around them) are equal. In fact, not only are the voltages around the parallel circuit elements equal to each other but they are the same as the source voltage, V_s. In other words:
V_s=V_1=V_2=V_3Therefore, the current equation can be rewritten as follows:
I=\frac{V_s}{R_1}+\frac{V_s}{R_2}+\frac{V_s}{R_3}=V_s\left\lgroup\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right\rgroupAnd the source current would be as follows:
I=V_s\left\lgroup\frac{R_1R_2+R_2R_3+R_1R_3}{R_1R_2R_3}\right\rgroupwhich is the same as the answer derived through Approach 1.
