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Step-by-Step Solution
Step 1: Identify the given information
• Magnetic energy stored in the coil, $U = 64\text{ J}$
• Power (rate of energy dissipation), $P = 640\text{ W}$
• Steady current flowing through the coil, $i = 8\text{ A}$
• We need to determine the time constant $\tau = \frac{L}{R}$ when the coil is connected to an ideal battery.
Step 2: Recall the formula for energy stored in an inductor
The magnetic energy $U$ stored in an inductor $L$ with current $i$ is given by:
$$U = \tfrac{1}{2} L\,i^2$$
Step 3: Calculate the inductance $L$
Given $U = 64\,\text{J}$ and $i = 8\,\text{A}$,
$$64 = \tfrac{1}{2} L \times (8)^2 \quad\Longrightarrow\quad 64 = \tfrac{1}{2} L \times 64$$
Solving for $L$:
$$64 = 32\, L \quad\Longrightarrow\quad L = 2\,\text{H}$$
Step 4: Use the power dissipation formula to find $R$
When a current $i$ passes through a resistor $R$, the power dissipated is:
$$P = i^2 R$$
Given $P = 640\,\text{W}$ and $i = 8\,\text{A}$,
$$640 = (8)^2 \, R$$
Solving for $R$:
$$640 = 64\, R \quad\Longrightarrow\quad R = 10\,\Omega$$
Step 5: Compute the time constant $\tau$
For an LR circuit, the time constant is:
$$\tau = \frac{L}{R}$$
Substitute $L = 2\,\text{H}$ and $R = 10\,\Omega$:
$$\tau = \frac{2}{10} = 0.2\,\text{s}$$
Final Answer
The time constant of the circuit is $0.2\,\text{s}$.
