Direct-Current Circuits11318.17 We name the currents as shown. Using Kirchhoff’s loop rule on the rightmost loop gives 123, , and III( )()325.001.00II−+12.0 V- 1.00+3.004.00 V0+−=(or )324.00 V=2.003.00+((1) Applying the loop rule to the leftmost loop yields ) ( )214.00 V+ 1.000(+5.008.00+−=or ) ( )2.00 V=4.003.00−123+=I(2) From Kirchhoff’s junction rule, (3) Solving equations (1), (2) and (3) simultaneously gives 3=0.846 A, 1.31 AI==0.462 A, and All currents are in the directions indicated by the arrows in the circuit diagram. ±²³³´±±µµ¶²³·¸²³³·µ²³³±±¶±±¹²³³±µ²³³±º²³³±18.18 Observe that the center branch of this circuit, that is the branch containing points aand b, is not a continuous conducting path, so no current can flow in this branch. The only
This is the end of the preview. Sign up
access the rest of the document.