Electromagnetic Induction And Alternating Current Ques 51

Answer:

Correct Answer: 51.$3 \times 10^{-5}, 10$

Solution:

Formula:

Equivalent self inductance :

51. Inductance of the circuit $L=\frac{0.9 \times 10^{-4}}{2}=0.45 \times 10^{-4} \mathrm{H}$ (in parallel)

Resistance of the circuit $R=3 / 2=1.5 \Omega$ (in parallel)

$\therefore \quad \tau_{L}($ time constant $)=\frac{L}{R}=3.0 \times 10^{-5} \mathrm{~s}$

Steady state current in the circuit through the battery

$$ i_{0}=\frac{V}{R}=\frac{15}{1.5}=10 \mathrm{~A} $$