2. Identify the unknowns. You will need as manyindependent equations as there are unknowns. Youmay write down more equations than this, but youwill find that some of the equations will be redundant(that is, not be independent in the sense of providingnew information). You may usefor eachresistor, which sometimes will reduce the numberof unknowns. 3. Apply Kirchhoff’s junction rule at one or more junctions. V = IR I 1 , I 2 , I 3 h a b c d g f e 30 Ω 40 Ω 20 Ω r = 1 Ω e 1 = 80 V r = 1 Ω e 2 = 45 V I 3 I 1 I 2 FIGURE 19–13 Currents can be calculated using Kirchhoff’s rules. See Example 19 – 8. Using Kirchhoff’s rules. Calculate the currents and in the three branches of the circuit in Fig. 19 – 13 (which is the same as Fig. 19 – 11). APPROACH and SOLUTION 1. Label the currents and their directions. Figure 19 – 13 uses the labels and for the current in the three separate branches. Since (positive) current tends to move away from the positive terminal of a battery, we choose and to have the directions shown in Fig. 19 – 13. The direction of is not obvious in advance, so we arbitrarily chose the direction indicated. If the current actually flows in the opposite direction, our answer will have a negative sign. 2.Identify the unknowns. We have three unknowns and thereforewe need three equations, which we get by applying Kirchhoff’s junction andloop rules. 3. Junction rule : We apply Kirchhoff’s junction rule to the currents at point a, where enters and and leave: (i) This same equation holds at point d, so we get no new information by writing an equation for point d. I 3 = I 1 + I 2 . I 1 I 2 I 3 (I 1 , I 2 , and I 3 ) I 1 I 3 I 2 I 3 I 1 , I 2 , I 3 I 1 , I 2 , EXAMPLE 19 ; 8 P R O B L E M S O L V I N G Choose current directions arbitrarily
SECTION 19 – 3 Kirchhoff’s Rules 535 P R O B L E M S O L V I N G Be consistent w ith signs w hen applying the loop rule P R O B L E M S O L V I N G I 1 is in the opposite direction from that assumed in Fig. 19 – 13 4. Loop rule : We apply Kirchhoff’s loop rule to two different closed loops. First we apply it to the upper loop ahdcba. We start (and end) at point a. From a to h we have a potential decrease From h to d there is no change, but from d to c the potential increases by 45 V: that is, From c to a the potential decreases through the two resistances by an amount Thus we have or (ii) where we have omitted the units (volts and amps) so we can more easily see the algebra. For our second loop, we take the outer loop ahdefga. (We could have chosen the lower loop abcdefga instead.) Again we start at point a, and going to point h we have Next, But when we take our positive test charge from d to e, it actually is going uphill, against the current — or at least against the assumed direction of the current, which is what counts in this calculation. Thus has a positi v e sign.
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- Spring '14