Thus far, we have seen how the sum of voltage drops across a series circuit is equal to the voltage value of the source ( Vs ):
Vs = V1 + V2 + V3
In the aforementioned scenario, three resistors are situated within a non-diverging electrical path; thus, each resistor along the electrical path has the same current ( I ) run through it. This may be summed up in the following manner:
Vs = IRt
IRt = ( I )( R1 + R2 + R3 )
IRt = IR1 + IR2 + IR3
If we substitute the ohm’s law expression into the lattermost equation, we end up with a rudimentary form of the voltage-divider equation:
( Vs / Rt ) = I
IRt = IR1 + IR2 + IR3
( Vs / Rt )( Rt ) = ( Vs / Rt )( R1 ) + ( Vs / Rt )( R2 ) + ( Vs / Rt )( R3 )
Vs = ( R1 / Rt )( Vs ) + ( R2 / Rt )( Vs ) + ( R3 / Rt )( Vs )
Vs = ( Vs )[ ( R1 / Rt ) + ( R2 / Rt ) + ( R3 / Rt ) ]
Vs = ( Vs )[ ( R1 + R2 + R3 ) / Rt ]
Note: R1 + R2 + R3 = Rt
Each resistor contributes a fraction of the total resistance encountered by each coulomb of charge that traverses the circuit. As a consequence, the voltage drop across each resistor is proportional to its relative ability to resist the flow of current in the circuit. An equally useful approach can be made via mathematical evaluation of each resistor as a separate entity within the circuit:
( Vs / Rt ) = I
It = I1 = I2 = I3
I1 = ( V1 / R1 )
I2 = ( V2 / R2 )
I3 = ( V3 / R3 )
( Vs / Rt ) = ( V1 / R1 )
( Vs / Rt ) = ( V2 / R2 )
( Vs / Rt ) = ( V3 / R3 )
And,
( Vs )( R1 / Rt ) = V1
( Vs )( R2 / Rt ) = V2
( Vs )( R3 / Rt ) = V3
Once again, we see that the voltage drop across each resistor in a series circuit is proportional to how much it resists the flow of current. In addition to mathematical clarity, this observation makes intuitive sense as well. It takes more joules ( J ) of energy per coulomb ( C ) of charge to successfully run an electric motor as opposed to lighting a light bulb. The energy lost in each voltage drop is energy gained by something else:
Q: A 10 V battery is connected in series with two resistors. The first resistor has a value of 82 Ω, and the lattermost resistor has a value of 68 Ω. What are the voltage drops across each resistor?
A: ( Vs / Rt ) = ( V1 / R1 )
And,
( Vs / Rt ) = ( V2 / R2 )
( Vs )( R1 / Rt ) = V1
( Vs )( R2 / Rt ) = V2
Rt = R1 + R2
82 Ω + 68 Ω = 150 Ω
V1 = ( 10 V )( 82 Ω / 150 Ω )
V1 = ( 10 V )( 0.546 ) = 5.47 V
V2 = 10 V – 5.47 V = 4.53 V
Likewise, the second voltage value could have been determined via substitution:
V1 = ( 10 V )( 68 Ω / 150 Ω )
V1 = ( 10 V )( 0.453 )
V1 = 4.53 V
RENAISSANCE PHYSICS 