An emf of 10 V is applied to a coil with an inductance of 40 mH and a resistance

Question

An emf of 10 V is applied to a coil with an inductance of 40 mH and a resistance of 0.50 Ohm. How long does it take for the current to reach one half of its maximum value? What is the time constant? What is the voltage on the resistor after I time constant has elapsed? What is the current in the circuit after I time constant has elapsed? What is the current in the circuit when t = 0.04 s? What is the voltage on the resistor when t = 0.0 s? What is the steady state value of the current in the circuit? What is the steady state value of the voltage on the inductor?

Explanation / Answer

Here,

E = 10 V

L = 0.040 H

R = 0.50 Ohm

for the time taken for current to reach half value is t

Now , I = Io * (1 - e^(-t*R/L))

0.5 = (1 - e^(-t * 0.50/0.040))

solving for t

t = 0.0555 s

the time taken for current to reach half value is 0.0555 s

-----------------------------

Time constant = L/R

Time constant = 0.040/.50 s

Time constant = 0.08 s

-------------------------------

voltage on the resistor , t = T

V = E * (1 - e^(-t/T))

V = 10 * (1 - e^(-1))

V = 6.32 V

the voltage on the resistor is 6.32 V

-----------------------------------
after 1 time constant

I = E/R * (1 - e^(-t*R/L))

I = (10/0.5) * (1 - e^(-T/T))

I = 20 * (1 - e^-1)

I = 12.64 A

the current in the circuit after 1 time constant is 12.64 A

-----------------------------------------------
at t = 0.04 s

I = E/R * (1 - e^(-t*R/L))

I = 20 * (1 - e^(-0.04/.08 ))

I = 7.87 A

the current in the circuit at t = 0.04 s is 7.87 A

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