Question

A device is turned on, and 3.00 A flows through it 0.100 ms later. What is the self-inductance of the device if an induced 150 V emf opposes this?

a) 0.05 H
b) 0.15 H
c) 0.50 H
d) 1.50 H

Answer 1

Using the formula ℑ = -L( ΔI / Δt ), the self-inductance of the device is calculated to be 0.05 H after rounding, which means option (a) is correct.

Step-by-step explanation:

To answer the self-inductance question, we'll need to use the formula that relates the emf induced in an inductor to the change in current and the self-inductance:

ℑ = -L( ΔI / Δt )

Where:

• ℑ is the induced emf (in volts, V)
• ΔI is the change in current (in amperes, A)
• Δt is the change in time (in seconds, s)
• L is the self-inductance (in henrys, H)

We're given that ℑ = 150 V, ΔI = 3.00 A, and Δt = 0.100 ms which is 0.100 × 10-3 s.

Rearranging the formula to solve for L:

L = -ℑ / (ΔI / Δt)

Substituting the values we get:

L = -150 V / (3.00 A / (0.100 × 10-3 s))

L = -150 V / (3.00 A / 0.0001 s)

L = -150 V / (30000 A/s)

L = -0.005 H

The negative sign indicates the direction of the induced emf opposes the change in current, but the magnitude of self-inductance is 0.005 H, which corresponds to option (a) 0.05 H after rounding.