Power is the rate at which energy is deposited within ( or liberated from ) some medium. As pertains to electronics, the watt is a measure of how many joules ( J ) of energy are deposited per second within the resistive elements of a circuit. The SI unit of power is the watt ( W ). Even though the physical structure of series and parallel circuits differ, the concept of power within each circuit type is the same, as are the equations used to express it:
P = IV
P = I2R
P = V2 / R
Nonetheless, the mathematical techniques used to obtain current ( I ), voltage ( V ), and resistance ( R ) values will indeed differ. In series circuits, the sum of voltage drops across a circuit will equal the source voltage; contrarily, the voltage drops across parallel circuits are equal in magnitude.
The latter statement above is a common source of confusion amongst students studying introductory electronics, but this need not be the case. Parallel circuits are current dividers, and the magnitude of current flowing across any given parallel resistor may or may not be the same. A relatively high current deposits energy at a faster rate than its lower-current counterparts; however, each coulomb ( C ) of charge deposits an amount of energy that is proportional to the source voltage:
Q: Three resistors are situated parallel to one another, and their values are as follows:
R1 = 68 Ω
R2 = 33 Ω
R3 = 22 Ω
An unknown voltage source supplies 200 mA of current to the first node of the circuit. What is the total amount of power deposited within the system? What is the source voltage ( Vs )?
A: The net power can be determined via usage of the total resistance ( Rt ) and current ( It ) of the circuit. Our value should be lower than 22 Ω ( why? ):
Rt = 1 / [ ( 1 / R1 ) + ( 1 / R2 ) + ( 1 / R3 ) ]
Rt = 1 / [ ( 1 / 68 Ω ) + ( 1 / 33 Ω ) + ( 1 / 22 Ω ) ]
Rt = 11.1 Ω
Pt = It2Rt
Pt = ( 200 mA )2( 11.1 Ω )
Pt = 0.444 W = 444 mW
If we’d been given a voltage value, we could have determined the power drop across each individual resistor before adding them together to obtain Pt. Let’s now derive the value of the source voltage:
Vs = ItRt
Vs = ( 200 mA )( 11.1 Ω )
Vs = 2.22 V
Prior to determining the power across each individual resistor, let’s see if the power equations that contain voltage terms yield consistent results:
Pt = ItVs
Vs = ( Pt / It )
Vs = ( 444 mW / 200 mA ) = 2.22 V
Likewise,
Pt = Vs2 / Rt
Vs2 = PtRt
Vs = √ PtRt
Vs = √ [ ( 444 mW )( 11.1 Ω ) ]
Vs = 2.22 V
Alas, Mother Nature loves consistency, oooh!!! Now that the voltage-source value has been confirmed, we may solve for the power drop across each resistor:
P1 = Vs2 / R1
P1 = ( 2.22 V )2 / 68 Ω
P1 = 72.5 mW
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P2 = Vs2 / R2
P2 = ( 2.22 V )2 / 33 Ω
P2 = 149 mW
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P3 = Vs2 / R3
P3 = ( 2.22 V )2 / 22 Ω
P3 = 224 mW
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Pt = P1 + P2 + P3
Pt = 72.5 mW + 149 mW + 224 mW
Pt = 446 mW
This calculation additionally shows us how rounding to three significant digits can introduce uncertainty in our answers, however so slightly.
RENAISSANCE PHYSICS 