Direct-Current Circuits11518.21First simplify the circuit by combining the series resistors. Then, apply Kirchhoff’s junction rule at point ato find Next, we apply Kirchhoff’s loop rule to the rightmost loop to obtain 122.00 AII+=( ) ( )12.00−()8.00 V6.00−+=or ( )112.02.00I−1I=8.00 V6.00(A0−This yields Finally, apply Kirchhoff’s loop rule to the leftmost loop to obtain 11.78 AI=)( ) ( )110 A6.00I4.002.08.00 V0ε+−(−)(+=)or 14.002.00A6.001.78=+A8.00 V10.7 V−=18.22From Kirchhoff’s point rule, note that 1I2=−in the circuit shown at the right. Going counterclockwise around the upper pane of the circuit, Kirchhoff’s loop rule gives ( )( )20II−=2250 30 90 −Ω−Ω+Ωor 2170 90 Ω=Ω(1) Now, going counterclockwise around the outer perimeter
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.