We present examples of circuits with series and parallel resistors and the formulas to calculate the equivalent resistance of these groups of resistors.
Resistors in SeriesThe resistors R1, R2 ...., Rm in the circuit on the left side are said to be in series because the same current passes through them. They behave in the same way as the circuit on the right of resistance Req given by the sum of the resistances R1, R2 and R3.Req = R1 + R2 ....+ Rm  The current I in the above circuit is given byI = E / Req Resistors in ParallelThe voltage across each of the resistors R1, R2 ...., Rm in the circuit on the left is the same and therefore these resistors are said to be in parallel. They behave in the same way as the circuit on the right of resistance Req that is given by the equation: 1 / Req = 1 / R1 + 1 / R2 + .... + 1 / Rm 
Resistors in Series: Examples with Detailed SolutionsExample 1Find the current I passing through and the voltage across each of the resistors in the circuit below. 
The three resistor in series have a resistance Req given by the sum of the three resistances. Hence Req = 100 + 400 + 200 = 700 Ω The current I passing through R1, R3 and R3 is the same and is calculated as follows: I = 7 v / 700 Ω = 0.01 A The voltage across each resistance is calculated using Ohm's law as follows: The voltage across 100Ω: VR1 = 100 × I = 100 × 0.01 = 1 v The voltage across 400Ω: VR2 = 400 × I = 400 × 0.01 = 4 v The voltage across 200Ω: VR3 = 200 × I = 200 × 0.01 = 2 v Resistors in Parallel: Examples with Detailed SolutionsExample 2Find current I in the circuit below and the current passing through each of the resistors in the circuit.  Solution to Example 2The three resistors are in parallel and behave like a resistor with resistance Req given by 1 / Req = 1 / 100 + 1 / 400 + 1 / 200 Multiply all terms by 400 and simplify to obtain 400 / Req = 4 + 1 + 2 Solve for Req to obtain Req = 400 / 7 Ω The main current I is given by I = 7 / Req = 7 / (400 / 7) = 49 / 400 A We now use Ohm's law to find the current passing through each resistor. The current through the resistor of 100 Ω: I1 = 7 / 100 A The current through the resistor of 400 Ω: I2 = 7 / 400 A The current through the resistor of 200 Ω: I3 = 7 / 200 A As an exercise; check that the sum of the three currents above is equal to the current I = 49 / 400 A. Series and Parallel Resistors: Examples with Detailed SolutionsExample 3Find current I in the circuit below. 
Solution to Example 3The two resistors that are in series are grouped as Req1 in the equivalent circuit below and their resistance is given by the sum Req1 = 100 + 400 = 500 Ω The two resistors that are in parallel are grouped as Req2 in the equivalent circuit below and their resistance is given by the equation 1 / Req2 = 1 / 100 + 1 / 200 
Solve to obtain Req2 = 200 / 3 Ω Req1 and Req2 are in series and therefore are equivalent to R given by the sum R = Req1 + Req2 = 500 + 200 / 3 = 1700 / 3 Ω We now use Ohm's law to find current I. I = 6 / R = 6 / (1700 / 3) = 18 / 1700 A Example 4 Example 5 More References and LinksOhm's Law with ExamplesFree SAT II Physics Practice Solutions on DC Electric Circuits |
