L-R Circuits

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The circuit shown has been operating for a long time. The instant after the switch in the circuit labeled S is opened, what is the voltage (in volt) across the inductor $V_{L}$ ? Take $R_{1} = 4.0 Ω, R_{2} = 8.0 Ω and L = 2.5 H$

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Solution

$Current in the inductor before opening ' S^{'}$
$I = \frac{12}{(\frac{4 \times 8}{4 + 8})} = \frac{9}{2} = 4.5 A$
$\begin{array}{l} Since current in inductor does not change instantly, therefore, just after opening 'S', \\ \begin{array}{l} ∴ 12 + V_{L} - I R_{1} = 0 \\ \Rightarrow 12 + V_{L} - (4.5) (4) = 0 \\ {⇒V}_{L} = 6 V \end{array} \end{array}$

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L-R Circuits

Remember concepts with our Masterclasses.

The circuit shown has been operating for a long time. The instant after the switch in the circuit labeled S is opened, what is the voltage (in volt) across the inductor VL ? Take R1=4.0Ω,R2=8.0Ω and L=2.5H

Current in the inductor before opening ' S'I=124×84+8=92=4.5 ASince current in inductor does not change instantly, therefore, just after opening 'S',​∴12+VL−IR1=0⇒12+VL−4.54=0⇒VL=6V

Ready to Test Your Skills?

free study material

The circuit shown has been operating for a long time. The instant after the switch in the circuit labeled S is opened, what is the voltage (in volt) across the inductor $V_{L}$ ? Take $R_{1} = 4.0 Ω, R_{2} = 8.0 Ω and L = 2.5 H$