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Genrish500 [490]
3 years ago
14

When a battery, a resistor, a switch, and an inductor form a circuit and the switch is closed, the inductor acts to oppose the c

hange in the current.
How is the time constant of the circuit affected by doubling the resistance in the circuit?
Physics
1 answer:
vladimir1956 [14]3 years ago
5 0

Answer:Time constant gets doubled

Explanation:

Given

L-R circuit is given and suppose R and L is the resistance and inductance of the circuit then current is given by

i=i_0\left [ 1-e^{-\frac{t}{\tau }}\right ]

where i_0 is maximum current

i=current at any time

\tau =\frac{L}{R}=time\ constant

\tau '=\frac{2L}{R}=2\tau

thus if inductance is doubled then time constant also gets doubled or twice to its original value.                                      

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Canola oil is less dense than water, so it floats on water, but its index of refraction is 1.47, higher than that of water. When
kupik [55]

Answer:

therefore critical angle c= 69.79°

Explanation:

Canola oil is less dense than water, so it floats over water.

Given n_{canola}= 1.47

which is higher than that of water

refractive index of water n_{water}=1.33

to calculate critical angle of light going from the oil into water

we know that

sinc= \frac{n_{water}}{n_{canola}}

now putting values we get

sinc= \frac{1.33}{1.47}

c= sin^{-1}(\frac{1.33}{1.47} )

c=69.79°

therefore critical angle c= 69.79°

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