IAR1 – €1 + IB(R2 + R3) = 0 - -IĄR1 + €1 + IB(R2 + R3) = 0 IAR1 – €1 – Ic(R5 + R4) + €2 = 0 €2 – IĄRĮ + €1 – Ic(R5 + R4) = 0 IA = IB + Ic А -IB(R2 + R3) – Ic(R4 + R5) + €2 = 0
IAR1 – €1 + IB(R2 + R3) = 0 - -IĄR1 + €1 + IB(R2 + R3) = 0 IAR1 – €1 – Ic(R5 + R4) + €2 = 0 €2 – IĄRĮ + €1 – Ic(R5 + R4) = 0 IA = IB + Ic А -IB(R2 + R3) – Ic(R4 + R5) + €2 = 0
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter27: Direct-current Circuits
Section: Chapter Questions
Problem 22P: For the circuit shown in Figure P27.22, we wish to find the currents I1, I2, and I3. Use Kirchhoffs...
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