Q1:Wires made of same materialEnergy conservation (loop) eqn:1) +emf – E*(2L1 +L2) = 02) +emf + E*(2L1 +L2) = 03) +emf – 2E1L1 – E2L2 = 04) +emf + 2E1L1 – E2L2 = 05) None of the aboveQ2:Wires made of different materials, but same cross-sectional areasEnergy conservation (loop) eqn:1) +emf – E1L1 – E2L2 = 02) +emf + E1L1 + E2L2 = 03) +emf – E*(L1 +L2) = 04) +emf + E*(L1 +L2) = 05) None of the aboveQ3:Same length and cross-sectional area, but different materials.Same u's, but n1 = 2*n2 1) E2 = emf/(1.5*L)2) E2 = emf/L3) E2 = emf/(2*L)4) E2 = 1.5*emf/LQ1What is the pattern of electric field in this steady-state circuit?NOT USED THIS TIME; Quiz covers much of thisQ1:What is the pattern of electric field in this steady-state circuit?1 32 4Q2:What charges make the electric field inside the wire in this circuit?The moving electrons inside the wire Charges on the battery and the surface of the wire Only charges on the batteryOnly charges on the surface of the wireQ3:Circuit 1: 1 battery, NiCr wire length Lcross-sectional area Aelectric field E1 inside wireCircuit 2: 1 battery, NiCr wire length (3L)cross-sectional area Aelectric field E2 inside wire.Which statement is correct?E1 = E2E1 = 3*E2E1 = E2/3Q4:Circuit 1: 1 battery, NiCr wire length Lcross-sectional area Aelectric field E1 inside wireCircuit 2: 1 battery, NiCr wire length Lcross-sectional area (4A)electric field E2 inside wire.Which statement is correct?E1 = E2E1 = 4*E2E1 = E2/4Q5:Circuit 1: 1 battery, NiCr wire length Lcross-sectional area ACircuit 2: 1 battery, NiCr wire length Lcross-sectional area (4A)Which statement is correct? i1 = i2 i1 = 4*i2i1 =
View Full Document
Unlocking...