Question

The current (i) at time t=0 and t=∞ respectively for the given circuit is : 

• A. 18E/55 , 5E/18
• B. 5E/18 , 18E/55
• C. 10E/33 , 5E/18
• D. 5E/18 , 10E/33

Hint:

Inductors resist changes in current (open at $t=0$), while capacitors resist changes in voltage (short at $t=0$).

Answer 1

The Correct Option is D

Let's solve the given problem by analyzing the circuit at two points: at time \(t = 0\) and at \(t = \infty\).

1. At \(t = 0\):

Initially, the inductor behaves like a short circuit, so the circuit will be as follows:

• Calculate the equivalent resistance seen by the source:
• The two parallel branches (upper and lower) have resistances of \(6 \, \Omega\) each: 

The equivalent resistance \(R_{eq}\) for each branch is:

\(R_{eq} = \frac{5 \times 1}{5 + 1} = \frac{5}{6} \, \Omega\)

\(R_{eq} = \frac{5 \times 4}{5 + 4} = \frac{20}{9} \, \Omega\)

So, the total resistance is \(R_{T} = \frac{5}{6} + \frac{20}{9} = \frac{55}{18} \, \Omega\)

• The current \(i_0\) at \(t = 0\) is given by Ohm's Law:

\(i(0) = \frac{E}{R_{T}} = \frac{E}{\frac{55}{18}} = \frac{18E}{55}\)

1. At \(t = \infty\):

At steady state, the inductor behaves like an open circuit, so the lower branch is open. The current will only flow through the upper branch:

• The resistance of the remaining path is:

\(R_{ss} = 5 + 5 = 10 \, \Omega\)

• The current \(i_{\infty}\) through the circuit:

\(i(\infty) = \frac{E}{R_{ss}} = \frac{E}{10}\)

• The correct answer, therefore, is:

\(i(0) = \frac{18E}{55}, \; i(\infty) = \frac{10E}{33}\)

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