Answer:
0.4
Explanation:
Given that a particular inductor is connected to a circuit where it experiences a change in current of 0.8 amps every 0.10 sec. If the inductor has a self-inductance of 2.0 V, what is the inductance
Using the power formula
P = IV
Substitute all the parameters
P = 0.8 × 2
P = 1.6 W
But P = I^2 R
Substitute power and current
1.6 = 0.8^2 R
R = 1.6 / 0.64
R = 2.5 ohms
Inductance = reciprocal of resistance
Inductance = 1 / 2.5
Inductance = 0.4
B
paired electrons spin in opposite directions cancelling their magnetic fields
Answer: current I = 1.875A
Explanation:
If the resistors are connected in series,
Then the equivalent resistance will be
R = 6 + 18 + 15 + 9
R = 48 ohms
Using ohms law
V = IR
Make current I the subject of formula
I = V/R
I = 90/48
I = 1.875A
And if the resistors are connected in parallel, the equivalent resistance will be
1/R = 1/6 + 1/18 + 1/15 + 1/9
1/R = 0.166 + 0.055 + 0.066 + 0.111
R = 1/0.3999
R = 2.5 ohms
Using ohms law
V = IR
I = 90/2.5
Current I = 35.99A
