High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Physics Textbook (Master) > DC Circuits

Grounding in Circuits

RLC CircuitsPrevious Section
Basics of AC CircuitsNext Section
Popular Courses
Find My Course
0:00 / 0:00

Grounding in Circuits


When talking about electric potential, we always need a reference point (the zero or ground level).

Although the potential difference (between two points in space) is always the same irrespective of the reference point, we get different values for the electric potential depending on what we chose as our zero level.

Therefore, grounding is a useful way of defining this zero level for the potential, so that all other voltages in the circuit can be computed relative to it.















Wize Concept
The ground in this context is not a connection to the physical earth. It's just a point in the circuit we choose to be our reference, at which we define the potential to be V=0V=0.


Exam Tip
To find the potential at a given point in the circuit, go towards it along any path beginning at the grounding point, and add together all the voltage rises and drops along the way to get the voltage at the given point.

0:00 / 0:00

Example: Grounding


The circuit below is grounded at point A. What is the potential at points A, B, C and D in the circuit? We have R1=2R_1=2 Ω, R2=5R_2=5 Ω, R3=1R_3=1 Ω and R4=6R_4=6 Ω. The battery has a voltage of 1212 V.




The potential at point A is zero by definition, since that's where the circuit is grounded:

VA=0V_A=0


Let's find the equivalent resistance for the entire circuit:

Resistors 3 and 4 are in parallel so we get:

1R34=1R3+1R4=11+16=76   →   R34=67\dfrac{1}{R_{34}}=\dfrac{1}{R_3}+\dfrac{1}{R_4} = \dfrac{1}{1} + \dfrac{1}{6} = \dfrac{7}{6} \ \ \ \to \ \ \ R_{34} = \dfrac{6}{7}

This combined resistance is in series with resistors 1 and 2:

Req=R1+R2+R34=2+5+67=7.86R_{eq}=R_1+R_2+R_{34}=2+5+\dfrac{6}{7}=7.86 (Ω)

The current through the entire circuit is also the current through resistors 1 and 2, and is given by:

V=IReqV=IR_{eq}

I=VReq=127.86=1.53I=\dfrac{V}{R_{eq}}=\dfrac{12}{7.86}=1.53 (A)



To find the potential at point B, consider any path going from A to B, and add all the voltages along the way to get the potential at B.

The easiest one to choose is going clockwise from A to B:

VA=VBV_A=V_B

Since VA=0V_A=0 we get VB=0V_B=0 as well.


Do the same between points A and C, going clockwise, to get the potential at point C:

VA−IR1=VCV_A-IR_1=V_C

0−IR1=VC0-IR_1=V_C

VC=−IR1=−(1.53)(2)=−3.05V_C=-IR_1=-(1.53)(2)=-3.05 (V)


A clockwise path between points A and D gives us the potential at point D:

VA−IR1−IR2=VDV_A-IR_1-IR_2=V_D

0−IR1−IR2=VD0-IR_1-IR_2=V_D

VD=−IR1−IR2=−(1.53)(2)−(1.53)(5)=−10.69V_D=-IR_1-IR_2=-(1.53)(2)-(1.53)(5)=-10.69 (V)