B
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Answer:
Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.
Explanation:
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My guess is A. I'm not 100% positive but i'm pretty sure.
Complete question:
A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?
Answer:
The current in the circuit 7 ms later is 0.2499 A
Explanation:
Given;
Ideal inductor, L = 45-mH
Resistor, R = 60-Ω
Ideal voltage supply, V = 15-V
Initial current at t = 0 seconds:
I₀ = V/R
I₀ = 15/60 = 0.25 A
Time constant, is given as:
T = L/R
T = (45 x 10⁻³) / (60)
T = 7.5 x 10⁻⁴ s
Change in current with respect to time, is given as;

Current in the circuit after 7 ms later:
t = 7 ms = 7 x 10⁻³ s

Therefore, the current in the circuit 7 ms later is 0.2499 A
Answer:
C. Its velocity is perpendicular to its acceleration
Explanation:
Because acceleration is always perpendicular to the velocity when the velocity will change direction without change it's magnitude