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

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

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 and V = 0.8V₀.

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

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

nah.   i'm just sleepy.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft