In figure- final value of current in 10- resistor- when plug of key K is inserted

The correct option is D zeroAcross inductor Δ V = L didt in steady state didt=0 Hence, no voltage drop across the inductor in steady state i = VR=0 ...

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Standard XII

Physics

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In figure, fi...

Question

In figure, final value of current in $10 Ω$ resistor, when plug of key $K$ is inserted

\frac{3}{10} A

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\frac{3}{20} A

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\frac{3}{11} A

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zero

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Solution

The correct option is D zero
Across inductor
$Δ V = L \frac{d i}{d t}$
in steady state
$\frac{d i}{d t} = 0$
Hence, no voltage drop across the inductor in steady state
$i = \frac{V}{R} = \frac{0}{10} = 0 A$
which implies all the current passes through the inductor

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In figure, final value of current in 10Ω resistor, when plug of key K is inserted

The correct option is D zeroAcross inductorΔV=Ldidtin steady statedidt=0Hence, no voltage drop across the inductor in steady statei=VR=010=0Awhich implies all the current passes through the inductor

In figure, final value of current in $10 Ω$ resistor, when plug of key $K$ is inserted

The correct option is D zero
Across inductor
$Δ V = L \frac{d i}{d t}$
in steady state
$\frac{d i}{d t} = 0$
Hence, no voltage drop across the inductor in steady state
$i = \frac{V}{R} = \frac{0}{10} = 0 A$
which implies all the current passes through the inductor