1. FieldsCircuitsMI 1. Ia, IIa, IIIa part 1 of 2 bulb, Ezz be the electric field in the lower-left bulb, and E2p to be the electric field in the O 2. Ib, IIb, IIIb lower-right bulb. Choose the answer that identifies the cor- rect statements from the following list. (la) El = E2L 3. Ia, IIb, IIIa Ib) Ei > E2L 4. Ib, IIb, IIIa (Ila) E2L > E2R (IIb) E21 = E2R 5. Ib, IIa, IIIa (IIIa) Ezz + E2R = E (a) (IIIb) E2L + E2R > E; 6. Ib, IIa, IIIb 12 O 7. Ia, IIb, IIIb O 8. Ia, IIa, IIIb ibattery m (b) From a microscopic point of view, the im- portant quantities in a circuit analysis are the fields and the electron currents. This problem concerns the circuits shown in figures (a) and (b). The bulbs in both circuits are identical and have a filament length L, while the bat- teries are also identical with emf, E. Assume the potential difference along the connecting wires in both circuits is negligible. For the case of Fig(a), the field in the fila- ment is E. The circuit satisfies the loop equa- tion: & = EL and the electron current is Now consider the circuit of Fig(b). Define En to be the electric field through the topO 1. Ia, IIb, IIIb part 2 of 2 Denote the current through the top branch O 2. Ia, IIa, IIIa be i, and the current through the bottom branch to be i2. 3. Ib, IIa, IIIb Choose the answer that identifies the cor- rect statements from the following list. 4. Ia, IIa, IIIb (Ia) in = i (1b) in Ki 5. Ib, IIb, IIIb 6. Ib, IIa, IIIa (Ila) i2 = i O 7. Ib, IIb, IIIa (IIb) i2 = N/N. O 8. Ia, IIb, IIIa (IIIa) battery = (IIIb) ibattery = 2i