Question

A resistor, inductor, and a battery are arranged in a circuit. The circuit has an inductance of L = 1 H and a resistance of 1.4 kΩ. Switch S1 is suddenly closed at t = 0. Find the time needed for the current to reach a fraction f = 0.6 of its maximum value.

Answer 1

Answer:

$\tau \approx 7.14 \times 10^{-4}s \approx0.714ms$

Explanation:

In a LC circuit The time constant τ is the time necessary for 60% of the total current (maximum current), pass through the inductor after a direct voltage source has been connected to it. The time constant can be calculated as follows:

$\tau =\frac{L}{R}$

Therefore, the time needed for the current to reach a fraction f = 0.6(60%) of its maximum value is:

$\tau =\frac{1}{1.4\times 10^{3}} =7.142857143 \times 10^{-4} \approx7.14 \times 10^{-4}s$

Answer 2

The time needed for the current to reach a fraction of its maximum value will be "7.14 × 10⁻⁴ s".

Resistance and Current

According to the question,

Inductance, L = 1

Resistance, R = 1.4 kΩ

Fraction, f = 0.6 or.

                 = 60%

We know the relation,

→ Time constant, $\tau$ = $\frac{L}{R}$

or,

By substituting the values, we get

                               = $\frac{1}{1.4\times 10^3}$

                               = 7.142857143 × 10⁻⁴ or,

                               = 7.14 × 10⁻⁴ s

Thus the above response is correct.                                  

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