Question 201: AC fundamentals

Question

A coil has an inductance of $$2 \mathrm{H}$$ and resistance of $$4 ~\Omega$$. A $$10 \mathrm{~V}$$ is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be ___________ $$	imes 10^{-2} \mathrm{~J}$$.

Accepted Answer

To find the energy stored in the magnetic field after the current has built up to its equilibrium value, we first need to find the steady-state current in the coil.
When the current reaches its equilibrium value, the coil behaves like a resistor because the back-emf induced by the changing magnetic field is zero. Ohm's law can be applied:
$$I = \frac{V}{R}$$
where
$$I$$ is the current
$$V$$ is the voltage across the coil (10 V)
$$R$$ is the resistance of the coil (4 ÃÂ©)
Plugging in the values:
$$I = \frac{10}{4}$$
$$I = 2.5 \mathrm{~A}$$
Now that we have the steady-state current, we can find the energy stored in the magnetic field using the formula:
$$W = \frac{1}{2}LI^2$$
where
$$W$$ is the energy stored in the magnetic field
$$L$$ is the inductance of the coil (2 H)
$$I$$ is the steady-state current (2.5 A)
Plugging in the values:
$$W = \frac{1}{2}(2)(2.5)^2$$
$$W = 1(6.25)$$
$$W = 6.25 \mathrm{~J}$$
To express this in terms of $$10^{-2} \mathrm{~J}$$, divide by $$10^{-2}$$:
$$6.25 \div 10^{-2} = 625$$
Therefore, the energy stored in the magnetic field after the current has built up to its equilibrium value is 625$$	imes 10^{-2} \mathrm{~J}$$.

AC fundamentals

A coil has an inductance of $2 \mathrm{H} and resistance of 4 ~\Omega. A 10 \mathrm{~V} is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be ___________ \times 10^{-2} \mathrm{~J}$.

Integer / Numeric Answer Type