Variation of current in the circuit shown below needs to be studied as a function of the three resistors and the voltage source, V_s. Using Ohm’s law and KVL, develop an equation that can be used to compute the value of current I for various values of R_1, R_2, R_3, and V_s.
Similar to other engineering disciplines, in most electrical engineering problems, multiple methods can be employed to derive the solution. The suitability of one method over another depends on the known parameters and the complexity of the circuit.
APPROACH I
Reduce or simplify the given circuit to a “net” voltage source and equivalent resis -tance R_{eq}. Since R_1, R_2, and R_3 are in series:
R_{eq}=R_1+R_2+R_3
Based on the assumption that V_s is indeed the source driving this circuit, by electri-cal convention, the current in this circuit would flow “out” of the positive terminal, or anode, of the voltage source. Hence, the current would flow in the clockwise direction as shown in the diagram below.
By the same token, application of the electrical current convention to the voltage device, or load voltage, V_L would mean that it would try to set up current in the counterclockwise direction. However, because we assumed that voltage source V_s is driving the net flow of current through the circuit, its dominance over V_L is implied, and the net voltage in the circuit would be as follows:
V_{Net}=V_s-V_L
This results in the simplification of the circuit as depicted below:
Then, application of Ohm’s law yields:
I=\frac{(V_s-V_L)}{R_{eq}} or, I=\frac{V_s-V_L}{R_1+R_2+R_3}APPROACH II
This approach is premised on the application of KVL to the given circuit after the circuit has been annotated with voltage designations, voltage polarities, and cur-rent direction.
According to another electrical convention, voltage polarities are assigned such that the current enters the resistances (or loads, in general) on the positive side and exits from the negative side. The voltage sources or existing “voltage loads” retain their stated polarities. The aforementioned steps result in the transformation of the original (given) circuit as follows:
Apply Ohm’s law to define the voltages, or voltage drops, across the three resistors.
V_{R1}=IR_1 V_{R2}=IR_2 V_{R3}=IR_3With all voltages – voltage source, voltage load, and voltage drops across the resis-tors – identified and their polarities noted, apply KVL by “walking” the annotated circuit beginning at the cathode or negative electrode of the voltage source, V_s.
Note the voltages and respective polarities as you make a complete loop around the circuit in the clockwise direction of the current. This results in the following equation:
∑V=0-V_s+V_{R1}+V_L+V_{R2}+V_{R3}=0
Expansion of this KVL-based equation through substitution of the resistor voltage drop formulas, derived earlier, yields:
-V_s+I{R_1}+V_L+I{R_2}+I{R_3}=0Further rearrangement and simplification result in:
I(R_1+R_2+R_3)=(V_s-V_L)I=\frac{V_s-V_L}{R_1+R_2+R_3}