Below is a circuit schematic of sources and resistors (Figure 3). VS = 10V , R1 = 100Ω, R2 = 50Ω, R3 = 25Ω, IS = 2A. Calculate the voltage V3 and the current i1 in amps that goes through R3 from left to right.

Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.

So we first find the equivalent resistance for the two resistors in parallel:

[tex]\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega[/tex]

By knowing this, we can estimate the total current through the circuit,:

[tex]Vs=I\,*\,(\frac{100}{3} +25)\\10=I\,*\,\frac{175}{3} \\I=\frac{30}{175} \,amps[/tex]

So approximately 0.17 amps

and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

[tex]V_3=\frac{30}{175} *\,25=\frac{30}{7} \approx 4.28\,\,V[/tex]

So now we know that the potential drop across the parellel resistors must be:

10 V - 4.28 V = 5.72 V

and with this info, we can calculate the current through R1 using Ohm's Law:

[tex]I_1=\frac{V_1}{R_1} =\frac{5.72}{100} =0.0572\,\,amps[/tex]