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Kirchoff's Junction Law

In the given ...

Question

In the given branch, current in the branch is $i = 10 e^{- 4 t} A$ . If potential difference across inductor is $V_{L},$ potential difference between $A B$ is $V_{A} - V_{B},$ then

V_{L} = 80 e^{- 4 t} and V_{A} - V_{B} = 120 e^{- 4 t}

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V_{L} = 80 e^{4 t} and V_{A} - V_{B} = - 40 e^{- 4 t}

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V_{L} = 80 e^{4 t} and V_{A} - V_{B} = - 80 e^{- 4 t}

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V_{L} = 0 and V_{A} - V_{B} = - 120 e^{- 4 t}

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Solution

The correct option is A $V_{L} = 80 e^{- 4 t} and V_{A} - V_{B} = 120 e^{- 4 t}$
Given, $i = 10 e^{- 4 t}$

$V_{L} = ε = - L \frac{d i}{d t} = 80 e^{- 4 t}$

Using KVL to the open loop, we get,

$V_{A} - 4 i + L \frac{d i}{d t} = V_{B}$

$V_{A} - V_{B} = 4 i - L \frac{d i}{d t}$

$= 4 \times 10 e^{- 4 t} + V_{L}$

$= 40 e^{- 4 t} + 80 e^{- 4 t}$

$∴ V_{A} - V_{B} = 120 e^{- 4 t}$

 Hence, $(A)$ is the correct answer.

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