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The figure shows a circuit with two loops consisting of two horizontal branches and three vertical branches. The first horizontal branch has two resistors of 3 Ω each and the second branch has two voltage sources of 24 V with positive terminal on the left and 29 V with positive terminal on the right. The left vertical branch is directly connected, the middle branch has a resistance of 3 Ω and the right branch has a resistance of 4 Ω.
Choose the loops in the circuit.

Now we can apply Kirchhoff’s loop rule, using the map in [link] . Starting at point a and moving to point b , the resistor R 1 is crossed in the same direction as the current flow I 1 , so the potential drop I 1 R 1 is subtracted. Moving from point b to point e , the resistor R 2 is crossed in the same direction as the current flow I 2 so the potential drop I 2 R 2 is subtracted. Moving from point e to point f , the voltage source V 1 is crossed from the negative terminal to the positive terminal, so V 1 is added. There are no components between points f and a. The sum of the voltage differences must equal zero:

Loop a b e f a : I 1 R 1 I 2 R 2 + V 1 = 0 or V 1 = I 1 R 1 + I 2 R 2 .

Finally, we check loop ebcde . We start at point e and move to point b , crossing R 2 in the opposite direction as the current flow I 2 . The potential drop I 2 R 2 is added. Next, we cross R 3 and R 4 in the same direction as the current flow I 3 and subtract the potential drops I 3 R 3 and I 3 R 4 . Note that the current is the same through resistors R 3 and R 4 , because they are connected in series. Finally, the voltage source is crossed from the positive terminal to the negative terminal, and the voltage source V 2 is subtracted. The sum of these voltage differences equals zero and yields the loop equation

Loop e b c d e : I 2 R 2 I 3 ( R 3 + R 4 ) V 2 = 0 .

We now have three equations, which we can solve for the three unknowns.

(1) Junction b : I 1 I 2 I 3 = 0 . (2) Loop a b e f a : I 1 R 1 + I 2 R 2 = V 1 . (3) Loop e b c d e : I 2 R 2 I 3 ( R 3 + R 4 ) = V 2 .

To solve the three equations for the three unknown currents, start by eliminating current I 2 . First add Eq. (1) times R 2 to Eq. (2). The result is labeled as Eq. (4):

( R 1 + R 2 ) I 1 R 2 I 3 = V 1 . (4) 6 Ω I 1 3 Ω I 3 = 24 V .

Next, subtract Eq. (3) from Eq. (2). The result is labeled as Eq. (5):

I 1 R 1 + I 3 ( R 3 + R 4 ) = V 1 V 2 . (5) 3 Ω I 1 + 7 Ω I 3 = −5 V .

We can solve Eqs. (4) and (5) for current I 1 . Adding seven times Eq. (4) and three times Eq. (5) results in 51 Ω I 1 = 153 V , or I 1 = 3.00 A . Using Eq. (4) results in I 3 = −2.00 A . Finally, Eq. (1) yields I 2 = I 1 I 3 = 5.00 A . One way to check that the solutions are consistent is to check the power supplied by the voltage sources and the power dissipated by the resistors:

P in = I 1 V 1 + I 3 V 2 = 130 W , P out = I 1 2 R 1 + I 2 2 R 2 + I 3 2 R 3 + I 3 2 R 4 = 130 W .

Note that the solution for the current I 3 is negative. This is the correct answer, but suggests that the arrow originally drawn in the junction analysis is the direction opposite of conventional current flow. The power supplied by the second voltage source is 58 W and not −58 W.

Calculating current by using kirchhoff’s rules

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

The figure shows a circuit with three horizontal branches. The first branch has positive terminal of voltage source of 0.5 V connected to resistor R subscript 4 of 2 Ω, the second branch has negative terminal of voltage source of 0.6 V connected to resistor R subscript 3 of 1 Ω and the third branch has positive terminal of voltage source of 2.3 V connected to resistor R subscript 5 of 1 Ω. The left vertical branch has a resistor R subscript 1 of 3 Ω between first two horizontal branches and a resistor R subscript 2 of 5 Ω between the second and third horizontal branches. The right vertical branch is directly connected between first two horizontal branches and has a resistor R subscript 6 of 2 Ω between the second and third horizontal branches.
This circuit is combination of series and parallel configurations of resistors and voltage sources. This circuit cannot be analyzed using the techniques discussed in Electromotive Force but can be analyzed using Kirchhoff’s rules.

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 , I 2 , and I 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 apply the junction and loop rules, seeking three independent equations to allow us to solve for the three unknown currents.

Solution

Applying the junction and loop rules yields the following three equations. We have three unknowns, so three equations are required.

Junction c : I 1 + I 2 = I 3 . Loop a b c d e f a : I 1 ( R 1 + R 4 ) I 2 ( R 2 + R 5 + R 6 ) = V 1 V 3 . Loop c d e f c : I 2 ( R 2 + R 5 + R 6 ) + I 3 R 3 = V 2 + V 3 .

Simplify the equations by placing the unknowns on one side of the equations.

Junction c : I 1 + I 2 I 3 = 0. Loop a b c d e f a : I 1 ( 3 Ω ) I 2 ( 8 Ω ) = 0.5 V 2.30 V . Loop c d e f c : I 2 ( 8 Ω ) + I 3 ( 1 Ω ) = 0.6 V + 2.30 V .

Simplify the equations. The first loop equation can be simplified by dividing both sides by 3.00. The second loop equation can be simplified by dividing both sides by 6.00.

Junction c : I 1 + I 2 I 3 = 0. Loop a b c d e f a : I 1 ( 3 Ω ) I 2 ( 8 Ω ) = 1.8 V . Loop c d e f c : I 2 ( 8 Ω ) + I 3 ( 1 Ω ) = 2.9 V .

The results are

I 1 = 0.20 A , I 2 = 0.30 A , I 3 = 0.50 A .

Significance

A method to check the calculations is to compute the power dissipated by the resistors and the power supplied by the voltage sources:

P R 1 = I 1 2 R 1 = 0.04 W . P R 2 = I 2 2 R 2 = 0.45 W . P R 3 = I 3 2 R 3 = 0.25 W . P R 4 = I 1 2 R 4 = 0.08 W . P R 5 = I 2 2 R 5 = 0.09 W . P R 6 = I 2 2 R 6 = 0.18 W . P dissipated = 1.09 W . P source = I 1 V 1 + I 2 V 3 + I 3 V 2 = 0.10 W + 0.69 W + 0.30 W = 1.09 W .

The power supplied equals the power dissipated by the resistors.

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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