A uniformly wound solenoidal coil of self-inductance $1.8 \times 10^{- 4}$ H and resistance $6 Ω$ is cut into two identical coils. They are now connected in parallel across a $12$ volt battery of negligible resistance. Find the current drawn by the circuit in amp.

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Solution

The solenoid is cut into two equal half.

So the resistance of each part becomes $\frac{R}{2}$

Self inductance of each part also becomes $\frac{L}{2}$

Since these are connected in parallel

$L_{e q} = \frac{1}{\frac{1}{0.9 \times 10^{- 4}} + \frac{1}{0.9 \times 10^{- 4}}}$

$R_{e q} = \frac{1}{\frac{1}{3} + \frac{1}{3}}$

$R_{e q} = 1.5 Ω$

The current is given as,

$I = \frac{12 V}{1 .5 Ω}$

Hence the current is $10 A$

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A uniformly wound solenoidal coil of self-inductance 1.8×10−4 H and resistance 6Ω is cut into two identical coils. They are now connected in parallel across a 12 volt battery of negligible resistance. Find the current drawn by the circuit in amp.

The solenoid is cut into two equal half. So the resistance of each part becomesR2 Self inductance of each part also becomesL2 Since these are connected in parallel Leq=110.9×10−4+10.9×10−4 Req=113+13 Req=1.5Ω The current is given as, I=12V1.5Ω Hence the current is10A

A uniformly wound solenoidal coil of self-inductance $1.8 \times 10^{- 4}$ H and resistance $6 Ω$ is cut into two identical coils. They are now connected in parallel across a $12$ volt battery of negligible resistance. Find the current drawn by the circuit in amp.