The circuit can be reduced in steps as shown in Figure 21.20. The 8.0ohm and r =

Question

The circuit can be reduced in steps as shown in Figure 21.20. The 8.0ohm and r = 4.9 ohm resistors are in series, and so the equivalent resistance between a and b is x = 12.9 ohm (Eq. 21.27). The 6.0 ohm and 3.0 ohm resistors are in parallel, so from Equation 21.29 we find that the equivalent resistance from b to c is 2.0 ohm. Hence, what is the equivalent resistance from a to c? Req = 14.9 ohm Using Delta V = IR and the results from part A, we have the following. I= Delta V max/Req =48V/Req =3.22 A The current I in the 8.0 ohm and 4.9 ohm resistors is the same because the resistors are in series. At the junction at b, the current splits. Part of it (I1) is in the 6.0 Q resistor, and pert (I2) is in the 3.0 ohm resistor. Because the potential differences DeltaVbc across these resistors are the same (they are in parallel), we see that Delat V bc = IR =(6.0 ohm)I1 = (3.0ohm)I2, or I2 = 2.0I1. Using this result and that I1 +I2 =Ir we find the following. I1 = Your response differs from the correct answer by more than 10%. Double check your calculations. I2 = Your response differs from the correct answer by more than 10%. Ooubie check your calculations. We could have guessed this result by noting that the current in the 3.0 ohm resistor has to be twice the current in the 6.0 ohm resistor in view of their relative resistances and that the same potential difference appears across each of them. As a final check, note that Delta v dc =(6.0 ohm)I1 = (3.0 ohm)I2= 6.4 V and Delta V ab = (12.0 ohm)I = 41.6 V; therefore, Delata V ac = Delta V ab +DeltaVbc=48 V, as It must.

Explanation / Answer

net resistance = (3x6)/(3+6) + 8 + 4.9 ohm = 14.9 ohm total current = V/Rnet = 3.22 A now as current will split at the junction but the total current will be constant I1 + I2 = I I1 + 2 I1 = 3.22 3I1 = 3.22 I1 = 1.0733 A thus I2 = 2I1 = 2.1466A

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