A steady current of 2A flows in a coil of emf 12V for 0.4s.
A back emf of 3V was induced during this period.
The stored energy in the loop that can be utilized is?
Energy stored = Ivt = current x back e.m.f x time
= 2 x 3 x 0.4
= 2.4J
To calculate the stored energy in the coil, you can use the formula:
Energy = (current squared) x (inductance) x (time)
First, let's find the inductance of the coil. Inductance can be calculated using the formula:
Inductance = (emf - back emf) / current
Substituting the given values:
Inductance = (12V - 3V) / 2A
= 9V / 2A
= 4.5Ω
Next, let's calculate the energy:
Energy = (2A)^2 x (4.5Ω) x (0.4s)
= 4A^2 x 4.5Ω x 0.4s
= 7.2 Joules
Therefore, the stored energy in the loop that can be utilized is 7.2 Joules.
To calculate the stored energy in the loop that can be utilized, we need to calculate the energy stored in the coil when the current flows through it. The formula to calculate the energy stored in an inductor (coil) is:
Energy = (1/2) * L * I^2
Where:
- Energy is the stored energy in the inductor (coil),
- L is the inductance of the coil, and
- I is the current flowing through the coil.
Given:
- The current flowing through the coil (I) is 2A,
- The back emf induced (E) is 3V, and
- The time period (t) for which the current flows through the coil is 0.4s.
To calculate the inductance (L), we can use the formula:
L = (E * t) / I
Substituting the given values:
L = (3V * 0.4s) / 2A
L = 0.6Vs/A
Now, we can calculate the stored energy:
Energy = (1/2) * (0.6Vs/A) * (2A)^2
Energy = (1/2) * 0.6Vs * 4
Energy = 1.2 Joules
Therefore, the stored energy in the loop that can be utilized is 1.2 Joules.