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iris [78.8K]
3 years ago
10

A resistor, an inductor, and a switch are all connected in series to an ideal battery of constant terminal voltage. What does th

e time constant for the circuit represent? A resistor, an inductor, and a switch are all connected in series to an ideal battery of constant terminal voltage. What does the time constant for the circuit represent? The time constant represents the time required for the current to reach 25% of the maximum current. The time constant represents the time required for the current to reach 63% of the maximum current. The time constant represents the time required for the current to reach 99% of the maximum current. The time constant represents the time required for the current to reach 37% of the maximum current. The time constant represents the time required for the current to reach 75% of the maximum current. The time constant represents the time required for the current to reach 50% of the maximum current.
Physics
1 answer:
Irina-Kira [14]3 years ago
4 0

Answer:

The time constant τ = L/R represent the time requiered for the curent to get value of 63 % of its maximun value

Explanation:

In a circuit RL ( serie circuit with a resistor R and the inductor L ) when a voltage is applied the inductor will have a reaction, such reaction consist in the creation of an electromotive force which will prevent the current to get its maximun value. The time constant which in the case of an inductor is

τ = L/R ,  represent the time requiered for the crrent to be 63 % of its final value

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Answer:

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Explanation:

The formula for calculation an n^{th} bright fringe from the central maxima is given as:

y_n=\frac{n \lambda D}{d}

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y_2 = \frac{n \lambda_1 D}{d}

where;

n = 2

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substituting the parameters in the above equation; we have:

y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

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y_2 = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength \lambda_2 = 660-nm is as follows:

y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y^'}_2 = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

\delta y = y_2-y^{'}_2

\delta y = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

\delta y = 10 ⁻³ (2.76 - 1.94)

\delta y = 10 ⁻³ (0.82)

\delta y = 0.82 × 10 ⁻³ m

\delta y =  0.82 × 10 ⁻³ m (\frac{1.0mm}{10^{-3}m} )

\delta y = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

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