# How To Calculate Voltage Drop In A Series Circuit: Detailed Facts

This article elucidates what voltage drop is and How To Calculate Voltage Drop In A Series Circuit. Whenever the voltage meets any resistive element in the circuit, the value decreases or “drops.”

In a series circuit, there are several resistances or impedances. Each time when current passes through them, the voltage decreases. So, we need to know the value of a particular resistance and the current passing through it to compute the voltage drop across it. The voltage drop is current multiplied by the resistance.

## What is voltage drop?

Suppose we join a battery with a resistor through a wire. Electrons tend to flow from the negative to the positive side of the battery. It is like an electrical charge going from the positive to the negative terminal.

When one unit of charge encounters the resistor, it stops for a while. When it passes the resistor, another unit of charge comes and stops. At any time, the amount of charge at the end of the resistor is less than the charge at the start of the resistor. This phenomena creates a “potential or voltage drop.”

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## How to calculate total voltage drop in a series circuit?

Total voltage drop in a series circuit is the addition of all individual voltage drops caused by the impedance parameters. Also, the sum is equal to the total voltage supplied to the circuit or the voltage before any “drop.”

Let us examine the phenomena with the help of a circuit. In the circuit below, there are two resistors, R1 of 100 ohms and R2 of 200 ohms, connected with a supply V of 30 volt. Current [Latex] i = \frac {V} {R_{1} + R_{2}} = \frac {30} {100+200} = 0.1 \; A [/Latex]. Therefore voltage drops across R1 [Latex] = i \times R_{1} = 0.1 \times 100 = 10 \; V [/Latex] and across R2 [Latex] = i \times R_{2} = 0.1 \times 200 = 20 \; V [/Latex].

## How do you calculate voltage drop in an AC series circuit?

AC or alternating current circuits are electrical circuits with an AC supply voltage. An AC series circuit consists of any combination of resistor, inductor, and capacitor connected via series configuration.

Just like DC, we can compute the net impedance of an AC series circuit by adding them. The voltage drops can also be found in a similar manner. The voltage drop across any element in an AC series circuit is V= iZ, where Z is the net impedance of the circuit, and i is the total current flowing through it.

Read more on…..How To Calculate Voltage In A Series Circuit: Detailed Facts

## Voltage drop in series RLC circuit:

RLC circuit is a particular case of AC circuits. An RLC circuit comprises resistors, capacitors and inductors joined via series. Let us understand the voltage drops across an RLC series circuit through an example.

The circuit has three components drawn below: an R ohm resistor, an L Henry inductor, and a C farad capacitor. We have previously known, the voltage drop across any of them= impedance × current. So,

Voltage drop across the resistor = iR,  the inductor= iXL and the capacitor = iXC where XL= 2πfL and XC = 1/ 2πfC

## Q1. Three resistors are connected in series with values as R1= 4 Ω, R2= 5 Ω, and R3 = 6 Ω. The circuit is joined with a 15 V power supply. Find out the voltage drops across the resistors.

For calculating the potential drops across R1, R2, and R3, we first need to obtain the current in the circuit. We know, [Latex] current = \frac {net \; voltage} {equivalent \; resistance} [/Latex]

The equivalent resistance [Latex] R_{eq} = R_{1} + R_{2} + R_{3} = 4 + 5 + 6 = 15 \; \Omega [/Latex]

Therefore, total current [Latex] = \frac { 15 \: V} {15 \: \Omega } = 1\: A [/Latex]

Now, we can use ohm’s law (V = IR) for each resistor and find the voltage drops across them.

So, [Latex] V_{1} = I \times R_{1} = 1 \times 4 = 4 \; V [/Latex] ,

[Latex] V_{2} = I \times R_{2} = 1 \times 5 = 5 \; V [/Latex]

[Latex] V_{3} = I \times R_{3} = 1 \times 6 = 6 \; V [/Latex]

## Q2. For the below circuit, the voltage drop across the 6-ohm resistor is 12 V. Find out the other voltage drops and compute the total voltage drop or supply voltage.

We know, the voltage drop across any resistor in the series circuit = resistance × total current

If current i flows in the circuit, the voltage drop across the 6-ohm resistor is 6i.

6i = 12 or i = 2 amp

Therefore, voltage drop across the 2 ohm resistor = 2 x 2 = 4 V

Voltage drop across the 4 ohm resistor = 2 x 4 = 8 V

So the net voltage drop or the supply voltage = (12 + 4 + 8) = 24 V

## Q3. The image below depicts an RLC series circuit with the following components:A 120 V, 50 Hz AC supply, A 100-ohm resistor, A 20 μF capacitor, A 420 mH inductor.Calculate voltage drops across all three impedances.

We formerly knew how to calculate voltage drops for a series RLC circuit. The current multiplied by the impedance ( R or XL or XC) gives us the voltage drop. Let us find out XL and XC first.

XL= 2πfL (f is the frequency of the AC supply)

So, [Latex] X_{L} = 2 \times \pi \times 50 \times 420 \times 10^{-3} = 131.95 \;\; \Omega [/Latex]

[Latex] X_{C} = \frac {1} {2 \times \pi \times 50 \times 20 \times 10^{-6} }= 159.15 \;\; \Omega [/Latex]

Therefore, net impedance, [Latex] Z = \sqrt {(X_{C} – X_{L})^{2} + R^{2}} = \sqrt {(159.15-131.95)^{2} + 100^{2}} = 103.63\; \Omega [/Latex]

Now, for AC circuits, there is an entity called the phase angle. It gives a measure of the angle by which current lags or leads the voltage. Phase angle [Latex] \phi = \arctan (\frac{X_{C} – X_{L}} {R}) [/Latex]

[Latex] \phi = \arctan (\frac{27.2} {100}) = 15.22 ^{\circ} [/Latex]

So, current [Latex] i = \frac{ 120\angle 0} {103.63\angle 15.22} = 1.16\angle -15.22 ^{\circ}\; A [/Latex]

Therefore, [Latex] V_{R} = iR = 1.16\angle -15.22 \times 100\angle 0 = 116\angle -15.22\; V [/Latex]

Here, current leads the voltage as XC > XL.

[Latex] V_{C} = iX_{C} = 1.16\angle -15.22 \times 159.15\angle 90 = 184.61\angle 74.78\; V [/Latex]

[Latex] V_{L} = iX_{L} = 1.16\angle -15.22 \times 131.95\angle -90 = 153.06\angle -105.22\; V [/Latex]

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Kaushikee Banerjee

I am an electronics enthusiast and currently devoted towards the field of Electronics and Communications . My interest lies in exploring the cutting edge technologies. I'm an enthusiastic learner and I tinker around with open-source electronics. LinkedIn ID- https://www.linkedin.com/in/kaushikee-banerjee-538321175