INTRODUCTION TO ELECTRONICS: Parallel Circuits

In the study of parallel resistor ( R_x ) circuits, where “ x “ is the number of a particular resistor ( x = 1, 2, 3, … n ), a common point of confusion regards how the total resistance ( R_t ) of the circuit is always less than the lowest calculated resistor value. This statement is more of a linguistic riddle than anything else. To see why this is the case, we need to evaluate this statement from a conceptual vantage point. We will begin by looking at the formula for a parallel circuit’s total resistance:

R_t = 1 / [ ( 1 / R₁ ) + ( 1 / R₂ ) + ( 1 / R₃ ) + … ( 1 / R_n ) ]

The lowest resistor in such a circuit will have the largest quantity of current ( I ) passing through it; however, this quantity of current is always less than the total quantity of current traveling throughout the circuit’s branches. Let’s assume that the following current ( I_x ) value is the highest current traveling through the lowest of resistances :

I_t > I_x

( V / R ) = I

Since the voltage drops across a parallel circuit are equal to the source voltage ( V_s ), the expression above takes on the following form:

( V_s / R_t ) > ( V_s / R_x )

and,

R_t < R_x

Q: A voltage source of unknown magnitude gives rise to a 10 mA current as shown:

The current splits apart at a node and travels through three parallel resistances prior to recombining and returning to the source:

R₁ = 680 Ω

R₂ = 330 Ω

R₃ = 220 Ω

What is the magnitude of current traveling across each resistor?

A: This circumstance warrants the use of the current-divider formula:

I_x = ( R_t / R_x )( I_t )

Since we have been given the total current value and the value of each individual resistor, we must determine the total resistance value to solve the problem:

R_t = 1 / [ ( 1 / R₁ ) + ( 1 / R₂ ) + ( 1 / R₃ ) ]

R_t = 1 / [ ( 1 / 680 Ω ) + ( 1 / 330 Ω ) + ( 1 / 220 Ω ) ]

R_t = 111 Ω

I₁ = ( 111 Ω / 680 Ω )( 10 mA )

I₁ = ( 0.163 )( 10 mA )

I₁ = 1.63 mA

—————————————–

I₂ = ( 111 Ω / 330 Ω )( 10 mA )

I₁ = ( 0.336 )( 10 mA )

I₁ = 3.36 mA

——————————————

I₃ = ( 111 Ω / 220 Ω )( 10 mA )

I₃ = ( 0.505 )( 10 mA )

I₃ = 5.05 mA