INTRODUCTION TO ELECTRONICS: Two-Resistor Current Divider Derivation

We have previously seen how parallel circuits with two resistors ( R ) may be added together using specialized techniques. In one such case, the resistors had equal values, and in other cases, the values differed. These resistance values were then used to ascertain how currents ( I ) divide along each branch of the circuit in question. We will now develop derivations that justify usage of the following equations when warranted by appropriate circumstances:

R_p = R / n

I₁ = [ R₂ / ( R₁ + R₂ ) ]( I_t )

I₂ = [ R₁ / ( R₁ + R₂ ) ]( I_t )

The utility of these equations becomes apparent when the voltage ( V ) values across a two-resistor circuit are unknown. Let’s first consider a circumstance in which two ( and subsequently three ) 100 Ω resistors are added together:

Note: The three-resistor derivation will show that when resistor values are equal, the ( R / n ) derivation need not be restricted to two-resistor circuits.

R_t = 1 / [ ( 1 / 100 Ω ) + ( 1 / 100 Ω ) ]

R_t = 1 / [ ( 100 Ω + 100 Ω ) / ( 100 Ω )² ]

R_t = [ ( 100 Ω )² / 200 Ω ]

R_t = 50 Ω

Using the appropriate two-resistor formula, we obtain the same result:

R_t = R / n 

R_t = 100 Ω / 2 = 50 Ω

Since voltage drops across parallel circuits are equal, we may draw the following conclusions about current derivations in such a system:

( V_s / R₁ ) = I₁

( V_s / R₂ ) = I₂

Regardless of what the voltage value is, each resistor will divide into it equally; thus, the currents are equally split between the two branches in question. Let’s now see if the same consistency of results exists when three equal-value resistors are parallel to one another:

R_t = 1 / [ ( 1 / 100 Ω ) + ( 1 / 100 Ω ) + ( 1 / 100 Ω ) ]

R_t = [ 1 / ( 3 / 100 Ω ) ]

R_t = 100 Ω / 3 ≅ 33 Ω

Thus, for any number ( n ) of equal-value resistors within a parallel circuit, the total resistance is one-nth of the value of the resistors that are present. Of course, this will not be the scenario when resistance values differ:

V_s = V₁ = V₂

I₁R₁ = I₂R₂

I₁ = ( I₂ )( R₂ / R₁ )

I₁ = ( I₂ )( R₂ / R₁ )

Please recall that the total current in parallel circuits is a sum of the currents that travel through each branch:

I_t = I₁ + I₂ 

Let’s now substitute the value of I₁ above with the I₁ derivation prior to it:

I_t = [ ( I₂ )( R₂ / R₁ ) ] + I₂ 

I_t = ( I₂ )[ ( R₂ / R₁ ) ] + 1 ]

I_t = ( I₂ )[ ( R₂ + R₁ ) / R₁ ]

I₂ = ( I_t )[ R₁ / ( R₂ + R₁ ) ]

This result makes intuitive sense as well. If the resistance across R₁ is relatively small in comparison to the resistance across R₂, the current across R₂ will be relatively small, because the current across R₁ will be larger! It may first seem counterintuitive to use the resistance value of R₁ to determine the I₂ current value across R₂, but mathematics and intuition let us know that this must be the case. Let’s now continue:

I_t = I₁ + I₂ 

I₁ = I_t – I₂ 

I₁ = I_t – [ ( I_t )[ R₁ / ( R₂ + R₁ ) ]

I₁ = ( I_t )[ 1 – [ R₁ / ( R₂ + R₁ ) ]

I₁ = ( I_t )[ ( R₂ + R₁ ) / ( R₂ + R₁ ) – ( R₁ / ( R₂ + R₁ ) ]

I₁ = ( I_t )[ ( R₂ + R₁ – R₁ ) / ( R₂ + R₁ ) ]

I₁ = ( I_t )[ ( R₂  ) / ( R₂ + R₁ ) ]

Example: Determine the value of the current across I₄ in the diagram below where V_s = 5.0 V:

The total current ( I_t ) flowing through I₃ and I₄ is I₂. If we combine resistors R₃ and R₄, we can then add this equivalent resistance to R₂. Then, after deriving a single resistor ( R_t ) value across the second branch, we use ohm’s law to solve for I₂:

R_3∥4 = [ R₃R₄ / ( R₃ + R₄ ) ]

R_3∥4 = [ ( 330 Ω )( 560 Ω ) / ( 890 Ω ) ]

R_3∥4 = 208 Ω

R₂ + 208 Ω = 538 Ω

V_s / R = I

I_t = I₂ = ( 5.0 V / 538 Ω ) = 0.009293 = 9.29 mA

I₄ = [ R₃ / ( R₃ + R₄ ) ]( I_t )

I₄ = [ 330 Ω / ( 330 Ω + 560 Ω ) ]( 9.29 mA )

I₄ = ( 0.371 )( 9.29 mA )

I₄ = 3.45 mA