The resistor potential is constant and the inductor emf increases. The resistor potential is constant and the inductor emf decre
ases. The resistor potential is constant and the inductor emf is constant. The resistor potential decreases and the inductor emf is constant. The resistor potential decreases and the inductor emf increases. The resistor potential decreases and the inductor emf decreases. The resistor potential increases and the inductor emf increases. The resistor potential increases and the inductor emf is constant. The resistor potential increases and the inductor emf decreases.
1 answer:
The question is incomplete, the complete question is
When an RL circuit is connected to a battery, what happens to the potential difference across the resistor and the emf across the inductor?
The resistor potential is constant and the inductor emf increases.
The resistor potential is constant and the inductor emf decreases.
The resistor potential is constant and the inductor emf is constant.
The resistor potential decreases and the inductor emf is constant.
The resistor potential decreases and the inductor emf increases.
The resistor potential decreases and the inductor emf decreases.
The resistor potential increases and the inductor emf increases.
The resistor potential increases and the inductor emf is constant.
The resistor potential increases and the inductor emf decreases.
Answer:
The resistor potential increases and the inductor emf decreases.
Explanation:
From Kirchoff's rule, we can easily see that voltage across the inductor decreases steadily in a RL circuit until it finally gets to zero when the circuit is connected to a battery. A graph of the drop in potential across the inductor is attached for more clarity.
The drop shown in figure (b) in the image attached is the drop in potential across the inductor when an RL circuit is connected to a battery.
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Answer:
The earthquake occurred at a distance of 1122 km
Explanation:
Given;
speed of the P wave, v₁ = 8.5 km/s
speed of the S wave, v₂ = 5.5 km/s
The distance traveled by both waves is the same and it is given as;
Δx = v₁t₁ = v₂t₂
let the time taken by the wave with greater speed = t₁
then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.
v₁t₁ = v₂t₂
v₁t₁ = v₂(t₁ + 1.2 min)
v₁t₁ = v₂(t₁ + 72 s)
v₁t₁ = v₂t₁ + 72v₂
v₁t₁ - v₂t₁ = 72v₂
t₁(v₁ - v₂) = 72v₂
The distance traveled is given by;
Δx = v₁t₁
Δx = (8.5)(132)
Δx = 1122 km
Therefore, the earthquake occurred at a distance of 1122 km
Answer:
To obtain the power, we first need to find the work made by the force.
1) To calculate the work, we need the next equation:
So the force is given by the problem so our mission is to find 'dx' in terms of 't'
2) we know that:
So we have:
Then:
3) Finally, we replace everything:
After some calculation, we have as a result that the work is:
161.9638 J.
4) To calculate the power we need the next equation:
So
P = 161.9638/4.7 = 34.46 W