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mezya [45]
2 years ago
7

A series LR circuit contains an emf source of having no internal resistance, a resistor, a inductor having no appreciable resist

ance, and a switch. If the emf across the inductor is of its maximum value after the switch is closed, what is the resistance of the resistor
Physics
1 answer:
Natali5045456 [20]2 years ago
6 0

Answer:

b. 1.9 Ω

Explanation:

Here is the complete question

A series LR circuit contains an emf source of 14 V having no internal resistance, a resistor, a 34 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4.0 s after the switch is closed, what is the resistance of the resistor? a. 1.5 ? b. 1.9 ? c. 5.0 ? d. 14 ?

Solution

The voltage across the inductor V is

V = V_{0}e^{-\frac{Rt}{L} } where V₀ = emf of source = 14 V, R = resistance, L = inductance = 34 H and t = time

Given that V = 80% of its maximum value after 4.0 s, this implies that V = 80 % of V₀ = 0.8V₀ and t = 4.0 s

Since V = V_{0}e^{-\frac{Rt}{L} } and V = 0.8V₀.

Since we need to find R, we make R subject of the formula, we have

V = V_{0}e^{-\frac{Rt}{L} }

V/V_{0}= e^{-\frac{Rt}{L} }

taking natural logarithm of both sides, we have

㏑(V/V₀) = -Rt/L

R = -L㏑(V/V₀)/t

Substituting the values of the variables into the equation, we have

R = -34㏑(0.8V₀/V₀)/4.0 s

R = -34㏑(0.8)/4.0 s

R = -34 × -0.2231/4.0 s

R = 7.587/4

R = 1.896 Ω

R ≅ 1.9 Ω

So, B is the answer

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Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

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