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Although we know V5 = 7 V this equation leaves us with two unknowns. , With one equation and two unknowns we need more information to solve the problem. Some extra information comes in the form of Ohm s law. The same current I3 ows through the 10 and 3 resistors. Hence V10 = 10I3 and V3 = 3I3 and we can write (2.12) as 7 + 10I3 + 20 + 3I3 = 0 I3 = 1 A Knowing two of the currents, we can solve for the other current by considering KCL at the top-center node. We take + for currents entering the node and for currents leaving the node. This gives I1 I2 I3 = 0 Therefore we have I1 = I2 + I3 = 1.4 1 = 0.4 A Now we apply Ohm s law again to get the voltages across each of the resistors V10 = 10I3 = 10 ( 1) = 10 V V3 = 3I3 = 3 ( 1) = 3 V Notice the minus signs. These tell us that the actual voltages have the polarity opposite to that indicated in Fig. 2-8. EXAMPLE 2-6 Let s consider a simple abstract model of a toaster. Our model will consist of the wall outlet, an electric chord, a switch, and a heating element. The wall outlet
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then Exp3 is evaluated and its value becomes the value of the expression. Consider this example:
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But this last expression is a Riemann sum for the integral
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