Electromagnetic Induction and Alternating Current 4 Question 15
####15. A uniformly wound solenoidal coil of self-inductance $1.8 \times 10^{-4} \mathrm{H}$ and resistance $6 \Omega$ is broken up into two identical coils. These identical coils are then connected in parallel across a $15 \mathrm{~V}$ battery of negligible resistance. The time constant for the current in the circuit is …… s and the steady state current through the battery is … A. (1989, 2M)
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Answer:
Correct Answer: 15. $3 \times 10^{-5}, 10$
Solution:
- 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
$$ i_{0}=\frac{V}{R}=\frac{15}{1.5}=10 \mathrm{~A} $$
