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Calculating current: using kirchhoff’s rules

Find the currents flowing in the circuit in [link] .

The diagram shows a complex circuit with two voltage sources E sub one and E sub two and several resistive loads, wired in two loops and two junctions. Several points on the diagram are marked with letters a through h. The current in each branch is labeled separately.
This circuit is similar to that in [link] , but the resistances and emfs are specified. (Each emf is denoted by script E.) The currents in each branch are labeled and assumed to move in the directions shown. This example uses Kirchhoff’s rules to find the currents.

Strategy

This circuit is sufficiently complex that the currents cannot be found using Ohm’s law and the series-parallel techniques—it is necessary to use Kirchhoff’s rules. Currents have been labeled I 1 size 12{I rSub { size 8{1} } } {} , I 2 size 12{I rSub { size 8{2} } } {} , and I 3 size 12{I rSub { size 8{3} } } {} in the figure and assumptions have been made about their directions. Locations on the diagram have been labeled with letters a through h. In the solution we will apply the junction and loop rules, seeking three independent equations to allow us to solve for the three unknown currents.

Solution

We begin by applying Kirchhoff’s first or junction rule at point a. This gives

I 1 = I 2 + I 3 , size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +I rSub { size 8{3} } } {}

since I 1 size 12{I rSub { size 8{1} } } {} flows into the junction, while I 2 size 12{I rSub { size 8{2} } } {} and I 3 size 12{I rSub { size 8{3} } } {} flow out. Applying the junction rule at e produces exactly the same equation, so that no new information is obtained. This is a single equation with three unknowns—three independent equations are needed, and so the loop rule must be applied.

Now we consider the loop abcdea. Going from a to b, we traverse R 2 size 12{R rSub { size 8{2} } } {} in the same (assumed) direction of the current I 2 size 12{I rSub { size 8{2} } } {} , and so the change in potential is I 2 R 2 size 12{ - I rSub { size 8{2} } R rSub { size 8{2} } } {} . Then going from b to c, we go from to +, so that the change in potential is + emf 1 size 12{+"emf" rSub { size 8{1} } } {} . Traversing the internal resistance r 1 size 12{r rSub { size 8{1} } } {} from c to d gives I 2 r 1 size 12{ - I rSub { size 8{2} } r rSub { size 8{1} } } {} . Completing the loop by going from d to a again traverses a resistor in the same direction as its current, giving a change in potential of I 1 R 1 size 12{ - I rSub { size 8{1} } R rSub { size 8{1} } } {} .

The loop rule states that the changes in potential sum to zero. Thus,

I 2 R 2 + emf 1 I 2 r 1 I 1 R 1 = I 2 ( R 2 + r 1 ) + emf 1 I 1 R 1 = 0 . size 12{ - I rSub { size 8{2} } R rSub { size 8{2} } +"emf" rSub { size 8{1} } - I rSub { size 8{2} } r rSub { size 8{1} } - I rSub { size 8{1} } R rSub { size 8{1} } = - I rSub { size 8{2} } \( R rSub { size 8{2} } +r rSub { size 8{1} } \) +"emf" rSub { size 8{1} } - I rSub { size 8{1} } R rSub { size 8{1} } =0} {}

Substituting values from the circuit diagram for the resistances and emf, and canceling the ampere unit gives

3 I 2 + 18 6 I 1 = 0 . size 12{ - 3I rSub { size 8{2} } +"18" - 6I rSub { size 8{1} } =0} {}

Now applying the loop rule to aefgha (we could have chosen abcdefgha as well) similarly gives

+ I 1 R 1 + I 3 R 3 + I 3 r 2 emf 2 = + I 1 R 1 + I 3 R 3 + r 2 emf 2 = 0 . size 12{+I rSub { size 8{1} } R rSub { size 8{1} } +I rSub { size 8{3} } R rSub { size 8{3} } +I rSub { size 8{3} } r rSub { size 8{2} } - "emf" rSub { size 8{2} } "=+"I rSub { size 8{1} } R rSub { size 8{1} } +I rSub { size 8{3} } left (R rSub { size 8{3} } +r rSub { size 8{2} } right ) - "emf" rSub { size 8{2} } =0} {}

Note that the signs are reversed compared with the other loop, because elements are traversed in the opposite direction. With values entered, this becomes

+ 6 I 1 + 2 I 3 45 = 0 . size 12{+6I rSub { size 8{1} } +2I rSub { size 8{3} } - "45"=0} {}

These three equations are sufficient to solve for the three unknown currents. First, solve the second equation for I 2 size 12{I rSub { size 8{2} } } {} :

I 2 = 6 2 I 1 . size 12{I rSub { size 8{2} } =6 - 2I rSub { size 8{1} } } {}

Now solve the third equation for I 3 size 12{I rSub { size 8{3} } } {} :

I 3 = 22 . 5 3 I 1 . size 12{I rSub { size 8{3} } ="22" "." 5 - 3I rSub { size 8{1} } } {}

Substituting these two new equations into the first one allows us to find a value for I 1 size 12{I rSub { size 8{1} } } {} :

I 1 = I 2 + I 3 = ( 6 2 I 1 ) + ( 22 . 5 3 I 1 ) = 28 . 5 5 I 1 . size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +I rSub { size 8{3} } = \( 6 - 2I rSub { size 8{1} } \) + \( "22" "." 5 - 3I rSub { size 8{1} } \) ="28" "." 5 - 5I rSub { size 8{1} } } {}

Combining terms gives

6 I 1 = 28 . 5, and size 12{6I rSub { size 8{1} } ="28" "." 5} {}
I 1 = 4 . 75 A . size 12{I rSub { size 8{1} } =4 "." "75"" A"} {}

Substituting this value for I 1 size 12{I rSub { size 8{1} } } {} back into the fourth equation gives

I 2 = 6 2 I 1 = 6 9.50 size 12{I rSub { size 8{2} } =6 - 2I rSub { size 8{1} } =6 - 9 "." "50"} {}
I 2 = 3 . 50 A . size 12{I rSub { size 8{2} } = - 3 "." "50"" A"} {}

The minus sign means I 2 size 12{I rSub { size 8{2} } } {} flows in the direction opposite to that assumed in [link] .

Finally, substituting the value for I 1 size 12{I rSub { size 8{1} } } {} into the fifth equation gives

I 3 = 22.5 3 I 1 = 22.5 14 . 25 size 12{I rSub { size 8{3} } ="22" "." 5 - 3I rSub { size 8{1} } ="22" "." 5 - "14" "." "25"} {}
I 3 = 8 . 25 A . size 12{I rSub { size 8{3} } =8 "." "25"" A"} {}

Discussion

Just as a check, we note that indeed I 1 = I 2 + I 3 size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +I rSub { size 8{3} } } {} . The results could also have been checked by entering all of the values into the equation for the abcdefgha loop.

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Source:  OpenStax, General physics ii phy2202ca. OpenStax CNX. Jul 05, 2013 Download for free at http://legacy.cnx.org/content/col11538/1.2
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