Q 1 :

In the given circuit the sliding contact is pulled outwards such that electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12Ω, the value of the current in the circuit will be ________ A.          [2025]



(3)

εLdIdtIR=0

123×(8)I×12=0

 I=3 A



Q 2 :

Three identical resistors with resistance R=12Ω and two identical inductors with self inductance L = 5 mH are connected to an ideal battery with emf of 12 V as shown in figure. The current through the battery long after the switch has been closed will be ________ A.                  [2023]



(3)

After long time an inductor behaves as a resistance-less path. So current through cell

I=12R/3=3 A    {R=12Ω}



Q 3 :

As per the given figure, if dIdt=-1 A/s then the value of VAB at this instant will be _______ V.                   [2023]



(30)

dIdt=-IAsec

VA-IR-LdIdt-12=VB

VA-2×12-6(-1)-12=VB

VA-VB=36-6=30 V



Q 4 :

In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. Ea2b J/s represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of ba will be _________.                            [2023]



(25)

E=12LI2

Rate of energy storing =dEdt=LIdIdt

Now we know for an R-L circuit

        I=ER(1-e-RLt)

So  dIdt=ELe-ttRL

        dEdt=E2R(1-e-ttRL)(e-tRL)

Time at which rate of power storing will be maximum, t=LRln2

So  dEdt=E2R(1-12)×12

  E24R=E2100=E22×50

a=2,  b=50

So  ba=25