104CHAPTER 18 Problem Solutions 18.1From ( )VIRr+∆=, the internal resistance is 9.00 V72.0 4.92 0.117 A=− Ω=ΩVrRI∆=−18.2(a) 1234.0 8.0 12 24 eqRR=++=+Ω+ Ω= Ω(b) The same current exists in all resistors in a series combination. 24 V1.0 A24 eqVIR∆== =Ω(c) If the three resistors were connected in parallel, 111111112.18 4.0 8.0 12 eqRRRR−−=++=++ΩΩΩResistors in parallel have the same potential difference across them, so 4I424 V6.0 A4.0 VR∆===Ω, 824 V3.0 A8.0 IΩ, and 12
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