When the current in the portion of the circuit shown below is 2.40 A and increas

Question

When the current in the portion of the circuit shown below is 2.40 A and increases at a rate of 0.410 A/s, the measured voltage is ?Vab = 8.95 V. When the current is 2.40 A and decreases at the rate of 0.410 A/s, the measured voltage is ?Vab = 5.20 V. Calculate the values of L and R.

L =
Your response differs from the correct answer by more than 10%. Double check your calculations. H R =
Your response differs from the correct answer by more than 10%. Double check your calculations. ? When the current in the portion of the circuit shown below is 2.40 A and increases at a rate of 0.410 A/s, the measured voltage is? Vab = 8.95 V. When the current is 2.40 A and decreases at the rate of 0.410 A/s, the measured voltage is? Vab = 5.20 V. Calculate the values of L and R.

Explanation / Answer

in first case di/dt=0.410 A/s

apply kvl in circuit

Vab=Ldi/dt+i*R

i=2.4 A
di/dt=0.41 A/s

Vab=8.95 v

8.95=L*0.41+2.4*R

8.95=0.41L+2.4R...(1)

in second case di/dt decreases di/dt=-0.41 A/s

Vab=5.20 v

i=2.4 A
Vab=Ldi/dt+i*R

5.2=-0.41L+2.4R(2)

add (1)+(2)

[8.95+5.2]=0+[2.4+2.4]R

R=14.15/4.8=2.948 ohms

and (1)-(2)

[8.95-5.2]=[0.41+0.41]L+0

L=4.573 H

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