Question 584: Current Electricity

Question

If a current flowing in a coil is reduced to half of its initial value, the relation between the new energy $$\left(E_2\right)$$ and the original energy $$\left(E_1\right)$$ stored in the coil will be

Accepted Answer

To understand the relationship between the new energy $$\left(E_2\right)$$ and the original energy $$\left(E_1\right)$$ stored in the coil when the current is reduced to half of its initial value, we need to consider the formula for the energy stored in an inductor. The energy stored in a coil (inductor) is given by:
$$E = \frac{1}{2} L I^2$$
where $$E$$ is the energy, $$L$$ is the inductance of the coil, and $$I$$ is the current flowing through it.
If the current is reduced to half of its initial value, let the new current be $$I_2 = \frac{I_1}{2}$$ where $$I_1$$ is the original current.
Substituting $$I_2$$ into the energy formula, we get the new energy $$E_2$$:
$$E_2 = \frac{1}{2} L \left(\frac{I_1}{2}\right)^2$$
which simplifies to:
$$E_2 = \frac{1}{2} L \frac{I_1^2}{4}$$
Therefore:
$$E_2 = \frac{1}{2} \cdot \frac{1}{4} L I_1^2$$
Simplifying further:
$$E_2 = \frac{1}{4} \cdot \frac{1}{2} L I_1^2$$
Since $$E_1 = \frac{1}{2} L I_1^2$$, we see that:
$$E_2 = \frac{E_1}{4}$$
Thus, the correct option is Option A:
$$E_2 = \frac{E_1}{4}$$

Answer

$$E_2=\frac{E_1}{4}$$

Answer

$$\mathrm{E}_2=\frac{\mathrm{E}_1}{2}$$

Answer

$$\mathrm{E}_2=\mathrm{E}_1$$

Answer

$$\mathrm{E}_2=4 \mathrm{E}_1$$

Current Electricity

If a current flowing in a coil is reduced to half of its initial value, the relation between the new energy $\left(E_2\right) and the original energy \left(E_1\right)$ stored in the coil will be

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