Part F- What is bc at an intermediate time when i = 0.150 A? In, suppose that E = 60.0 V, R = 240 , and L = 0.160 H. Initiallythere is no current in the circuit. Switch S₂ is left open, and switchS₁ is closed.Part AJust after S₁ is closed, what is the potential differences Vab?Express your answer in volts.Π ΑΣΦVabSubmitPart BRequest AnswerVbc?Just after S₁ is closed, what is the potential differences Vbc ?Express your answer in volts.VE ΑΣΦ?VV+EfamsRS₁0000L$₂ Part CA long time (many time constants) after S₁ is closed, what is Vab?Express your answer in volts.VE ΑΣΦVab=SubmitPart DVbc =SubmitPart EA long time (many time constants) after S₁ is closed, what is Vbc?Express your answer in volts.15. ΑΣΦRequest AnswerVab==Request AnswerWhat is Vab at an intermediate time when i = 0.150 A?Express your answer in volts.ΓΠ ΑΣΦ??V?VV

The given circuit isϵ=60.0 V.R=240 Ω.L=0.16 H.Initially there is no current in the circuit.Switch S2 left open and switch S1 is closed.We need to determineA. The potential difference Vab just after S1 is closed.B. The potential difference Vbc just after S1 is closed.C. The potential difference Vab a long time after S1 is closed.D. The potential difference Vbc a long time after S1 is closed.E. The potential difference Vab at an intermediate time when i=0.15 A.F. The potential difference Vbc at an intermediate time when i=0.15 A.The current through the inductor just before the switch S1 is closed, i(0−)=0, A.But the current through the inductor can't change instantaneously. i.eThe current through the inductor just after the switch S1 is closed, i(0+)=0, A.i.e inductor will be open circuit.According to Ohm's law, the potential difference Vab just after the swtch S1 is closed is ⇒Vab=Ri(0+) ;⇒Vab=R(0) ;⇒Vab=0, V.By writing KVL equation in the loop in part (A),⇒−ϵ+Vab+Vbc=0 ;⇒−60+0+Vbc=0 ;⇒Vbc=60 V.Just after the swtch S1 is closed, the potential difference Vbc is⇒Vbc=60 V.A long time after S1 is closed means, the circuit will reach steady state.At steady state, the inductor will be short circuit.i.eBy writing KVL equation in loop,⇒−ϵ+Vab=0 ;⇒−60+Vab=0 ;⇒Vab=60 V.A long time after S1 is closed, the potential difference Vab is ⇒Vab=60 V.A long time after S1 is closed, the potential difference Vbc is ⇒Vbc=0 V.[ ∵ the potential difference across a short circuit is zero volts ]The circuit isThe potential difference Vab at an intermediate time when i=0.15 A is⇒Vab=Ri ;[ ∵ Ohm's law ]⇒Vab=(240)(0.15) ;⇒Vab=36 V.By writing KVL equation in loop in the above circuit,⇒−ϵ+Vab+Vbc=0 ;⇒−60+36+Vbc=0 ;⇒Vbc=24 V.The potential difference Vbc at an intermediate time when i=0.15 A is⇒Vbc=24 V.A.Just after the swtch S1 is closed, the potential difference Vab is⇒Vab=0 V.B.Just after the swtch S1 is closed, the potential difference Vbc is⇒Vbc=60 V.C.A long time after S1 is closed, the potential difference Vab is ⇒Vab=60 V.D.A long time after S1 is closed, the potential difference Vbc is ⇒Vbc=0 V.E.The potential difference Vab at an intermediate time when i=0.15 A is⇒Vab=36 V.F.The potential difference Vbc at an intermediate time when i=0.15 A is⇒Vbc=24 V.