Resistors in Series and Parallel

In a series circuit you simply add all the individual resistances: R_total = R₁ + R₂ + R₃ + … The reason: the same current flows through all components, and the partial voltages add up to the total voltage by the loop rule. Example: 150 Ω and 50 Ω in series give R_total = 200 Ω; at 12 V a current I = 12/200 = 0.06 A flows through both. The voltage divides in proportion to the resistances: 9 V drops across the 150 Ω resistor and 3 V across the 50 Ω resistor, which is the principle of the voltage divider. The total resistance is always larger than the largest single resistance.

In parallel the reciprocals add: 1/R_total = 1/R₁ + 1/R₂. Do not forget to take the reciprocal at the end! Example: 100 Ω in parallel with 300 Ω gives 1/R_total = 1/100 + 1/300 = 4/300, so R_total = 75 Ω. For exactly two resistors the product form is quicker: R_total = R₁·R₂/(R₁+R₂) = 30,000/400 = 75 Ω. For n equal resistors simply R/n holds: two 60 Ω resistors in parallel give 30 Ω. The physical reason: every additional parallel branch offers the current another path, the total current rises and the total resistance falls.

Because every additional branch opens another current path. The voltage across all branches is the same, so each branch carries its own current, and the currents add at the node. More total current at the same voltage necessarily means a smaller total resistance by R = U/I. Intuitively: a second checkout in a supermarket speeds things up even if it works more slowly than the first. This yields the most important sanity check for exams: the parallel resistance must be smaller than the smallest individual resistance involved. If you get 75 Ω for 100 Ω parallel to 300 Ω, it fits; if you got 150 Ω, there would be an arithmetic error somewhere.

Work from the inside out and combine step by step. First find groups that are clearly purely parallel or purely in series, replace them by their equivalent resistance and redraw the simplified circuit. Repeat until only one total resistance remains. Example: if R₂ = 100 Ω is parallel to R₃ = 300 Ω (giving 75 Ω) and this group is in series with R₁ = 25 Ω, then R_total = 100 Ω. Then calculate backwards: first the total current via I = U/R_total, then partial voltages and currents with Ohm law at each level. Cleanly redrawing after every step prevents most mistakes.

For two reasons: first, the same voltage lies across every parallel branch, so every device gets the full 230 V no matter how many other devices are running. Second, the branches work independently: if you switch one device off or a lamp burns out, current keeps flowing in the other branches. In a series circuit, by contrast, a single switched-off device would break the whole circuit, and the voltage would divide among all devices; an old string of fairy lights shows exactly this behaviour. The price of the parallel circuit: with every added device the total current rises, which is why the 16 A fuse limits how much may run at once.