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

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The circuit can be reduced in steps as shown in Figure 21.20. The 8.0 ? and r = 3.4 ? resistors are in series, and so the equivalent resistance between a and b is x = 11.4 ? (Eq. 21.27). The 6.0 ? and 3.0 ? resistors are in parallel, so from Equation 21.29 we find that the equivalent resistance from b to c is 2.0 ?. Hence, what is the equivalent resistance from a to c? Req = ? Your response differs from the correct answer by more than 10%. Double check your calculations. Using ?V = IR and the results from part A, we have the following. A The current I in the 8.0 ? and 3.4 ? 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 ? resistor, and part (I2) is in the 3.0 ? resistor. Because the potential differences ?Vbc across these resistors are the same (they are in parallel), we see that ?Vbc = IR = (6.0 ?)I1 = (3.0 ?)I2, or I2 = 2.0I1. Using this result and that I1 + I2 = I, we find the following. I1 = A I2 = A We could have guessed this result by noting that the current in the 3.0 ? resistor has to be twice the current in the 6.0 ? resistor in view of their relative resistances and that the same potential difference appears across each of them. As a final check, note that ?Vbc = (6.0 ?)I1 = (3.0 ?)I2 = 7.3 V and ?Vab = (12.0 ?)I = 41.7 V; therefore, ?Vac = ?Vab + ?Vbc = 49 V, as it must.

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please post the figure for question as it is hard to visualize from text.

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