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Electrical circuits transmit current through a series of components—power sources, conductors, resistors, and switches. When the supply voltage is insufficient, a voltage drop occurs, potentially starving devices of the power they need. Knowing how to calculate this drop across resistors helps you diagnose and design reliable circuits.
At the heart of these calculations is Ohm’s Law, which links voltage (V), current (I), and resistance (R) with the simple equation V = IR. This relationship lets you predict how much voltage a resistor will consume in a given circuit.
Consider a series circuit powered by a 30‑V battery with two resistors: 4 Ω and 6 Ω. The total resistance is 10 Ω, so the current is I = 30 V ÷ 10 Ω = 3 A. The voltage drop across each resistor follows directly:
Notice the sum of the drops (12 V + 18 V) equals the supply voltage, confirming the calculation.
In a more detailed series analysis, you can examine potential at points before, between, and after each resistor. For the same 30‑V source, let point A be before resistor 1, point B between the two, and point C after resistor 2. The drop across resistor 1 is the difference between A and B. If A is 30 V and B is 20 V, the drop is 10 V, giving a current of 10 V ÷ 4 Ω = 2.5 A. The same current flows through resistor 2, so its drop is 2.5 A × 6 Ω = 15 V. Point C is then 5 V, matching the remaining supply voltage.
Parallel circuits behave differently: the same voltage appears across each branch, while currents divide. With a 30‑V battery and three parallel resistors, each resistor experiences the full 30 V drop, regardless of its resistance value.
These calculations form the backbone of circuit design and troubleshooting. By applying Ohm’s Law correctly, you can ensure devices receive the required voltage and avoid failures caused by hidden drops.
