# What exactly is "voltage drop" across a circuit element?

2,825

## Solution 1

If you have a simple circuit with resistors, the voltage at the source decreases throughout the circuit. At the positive terminal of a battery for example, you might have a voltage of 9 V, while the negative terminal can be taken as the ground potential 0V.

The voltage in between those terminals (that means through the circuit) decreases from the positive voltage 9V to the ground potential 0 V.

How does it decrease? We assume that we have perfect conductors as wires, which means that they have the same potential at every point, i.e. at the beginning and the end of the wire, the voltage is equal. The only points in a circuit where the voltage can decrease are the resistors in that case. In front of the resistor is a higher voltage, and behind it a lower voltage. The difference is the voltage drop. It corresponds to the amount of power dissipated in the resistor and is proportional to the resistance.

Here is the schematic of a simple circuit. As you can see, there are three resistors in series and a voltage source of 9 V.

A current of $$0.5 mA$$ flows through the circuit, and it flows through all circuit elements. We can calculate this current by finding the equivalent resistance of the circuit. For a series connection of three resistors, we just add their resistances: $$R = R_1 + R_2 + R_3 = 3 k\Omega + 10 k\Omega + 5 k\Omega = 18 k\Omega$$

From $$U = R I$$ we can find the current flowing through every element: $$I = \frac{U}{R} = \frac{9 V}{18 k\Omega} = 0.5 mA$$

Now we can find how much voltage drops at every element. It is proportional to its resistance. Since we now know the current and the resistance at each element, we can calculate the voltage drop.

For the voltage that falls off at $$R_1$$, we write: $$\Delta U_1 = R_1 I = 3 k\Omega \cdot 0.5 mA = 1.5 V$$

Similarly for the other resistors we get $$5 V$$ for $$R_2$$ and $$2.5 V$$ for $$R_3$$.

In the circuit above we can see how this changes the voltage levels at each part of the circuit. Coming from the positive terminal of the voltage source, the voltage is 9 V. It stays this until it reaches the first resistor. At the front of the resistors (in direction of the current, clockwise), the voltage is 9 V (the red part), but the voltage behind the resistor is lower: it is now only $$9 V - 1.5 V = 7.5 V$$ (the yellow part). The voltage dropped over the resistor. If you'd probe the voltage behind the resistor, you should measure 7.5 V.

Similarly this works for all subsequent resistors: the 10k resistor will drop 5 V, which means that in front of it we have a level of 7.5 V (yellow), and behind it $$7.5 V - 5 V = 2.5 V$$ (green). Finally, the last resistor drops 2.5 V, bringing the voltage at the end down to 0 V (blue).

## Solution 2

when electrical current flows through a circuit component (easy example: a resistor), the resistor dissipates power; as it does, the voltage decreases steadily from the high-voltage side to the low-voltage side. this decrease in voltage is called the voltage drop.

In the case of a resistor of resistance R with a voltage difference across it of V volts, the power dissipated is equal to (V^2)/R.

Share:
2,825

Author by

### ten1o

Updated on August 01, 2022