INTRODUCTION TO ELECTRONICS: Power in Series Circuits

Thus far, we have seen how the net resistance ( R ) to current ( I ) flow within a series circuit is the sum of all the resistors that are present:

R_t = R₁ + R₂ + R₃ +…R_n

The voltage ( V ) drop that occurs as a coulomb ( C ) of charge passes through each resistor is a product of the current shared by all circuit elements and the resistance value of each resistor:

V_t = IR_t

IR_t = ( I )( R₁ + R₂ + R₃ +…R_n )

IR_t = IR₁ + IR₂ + IR₃ +…IR_n 

V_t = V₁ + V₂ + V₃ +…V_n

The power ( P ) that is consumed by each resistor is a measure of the energy drop in joules ( J ) that occurs each second ( s ). The SI unit of power is the Watt ( W ):

P = J / s

This expression can be derived via the product of voltage and current:

( J / C )( C / s ) = J / s

P = IV

And thus, the total power consumed within a series circuit is as follows:

P_t = P₁ + P₂ + P₃ +…P_n 

Furthermore, we may substitute each variable in this equation with ohm’s law derivations:

P = ( I )( IR ) = I²R

And,

P = ( V / R )( V ) = V² / R

Q: What is the net quantity of power consumed by the following series circuit:

A:  R_t = R₁ + R₂ + R₃ + R₄  

R_t = 10Ω + 12Ω + 56Ω + 22Ω

R_t = 100Ω

( V_t / R_t ) = I

( 15 V / 100Ω ) = 0.15 A = 150 mA

P = I²R_t

P = ( 150 mA )²( 100Ω )

P = ( 2.25 x 10^-2 A² )( 100Ω ) = 2.25 W

Likewise,

P = IV

P = ( 150 mA )( 15 V ) = 2.25 W

And finally,

P = V² / R_t

P = [ ( 15 V )² / 100Ω ] = 2.25 W

Additionally, the net power lost is a sum of the power lost at each individual resistor:

P_t = P₁ + P₂ + P₃ + P₄ 

P₁ = ( 150 mA )²( 10Ω ) = 225 mW

P₂ = ( 150 mA )²( 12Ω ) = 270 mW

P₃ = ( 150 mA )²( 56Ω ) = 1.26 W

P₄ = ( 150 mA )²( 22Ω ) = 495 mW

P_t = 225 mW + 270 mW + 1,260 mW + 495 mW = 2.25 W