(1) I 1 + I 2 - I 3 = 0 (2) Loop abcda :11.3V - (6.0 ? ) I 1 - (2.0 ? ) I 3 = 0

Question

(1)I1+I2-I3= 0 (2) Loopabcda:11.3V - (6.0?)I1- (2.0?)I3= 0
(3) Loopbefcb: -14.3V -11.3V + (6.0?)I1- (4.0?)I2= 0 11.3- 6.0I1- 2.0(I1+I2) = 0
(4)11.3= 8.0I1+ 2.0I2 (5)-12.8= -3.0I1+ 2.0I2 Example 21.9 Applying Kirchhoff's Rules Find the currentsI1,I2, andI3in the circuit shown in Figure 21.22, withV1=14.3V andV2=11.3V.

Find the potential difference between pointsbandc.

StrategyUse the loop rule and the junction rule.
Figure 21.22A circuit that cannot be simplified by using the rules for series and parallel resistors.
Solution We choose the directions of the currents as in Figure 21.22. Applying Kirchhoff's first rule to junctioncgives the following. (1)I1+I2-I3= 0 There are three loops in the circuit:abcda, befcb, andaefda(the outer loop). We need only two loop equations to determine the unknown currents. The third loop equation would give no new information. Applying Kirchhoff's second rule to loopsabcdaandbefcband traversing these loops in the clockwise direction, we obtain the following expressions. (2) Loopabcda:11.3V - (6.0?)I1- (2.0?)I3= 0
(3) Loopbefcb: -14.3V -11.3V + (6.0?)I1- (4.0?)I2= 0 Note that in loopbefcb, a positive sign is obtained when traversing the 6.0?resistor because the direction of the path is opposite the direction ofI1. Loopaefdagives -14.3V - (2.0?)I3- (4.0?)I2= 0, which is just the sum of (2) and (3).

Expressions (1), (2), and (3) represent three independent equations with three unknowns. We can solve the problem as follows. Dropping the units for simplicity and substitutingI3from (1) into (2) gives the following. 11.3- 6.0I1- 2.0(I1+I2) = 0
(4)11.3= 8.0I1+ 2.0I2 Dividing each term in (3) by 2 and rearranging the equation gives the following. (5)-12.8= -3.0I1+ 2.0I2 Subtracting (5) from (4) eliminatesI2, giving the following. 24.1= 11.0I1
I1=A
Using this value in (5) gives a value forI2. I2=A
Finally,I3=I1+I2. I3=A
Your response differs from the correct answer by more than 10%. Double check your calculations. That some of the values are negative indicates only that we chose the wrong directions for those currents. The numerical values, however, are correct.

In traversing the path frombtocalong the central branch, we have the following. Vc- Vb=11.3V - (6.0?)I1=V
Your response differs from the correct answer by more than 10%. Double check your calculations.
You can practice applying Kirchhoff's rules further withInteractive Example 21.9.

Explanation / Answer

there is no fig??

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