Two circuits A and B are connected to identical d.c. sources each of e.m.f. 10 volt. Self-inductances are $L_A = 10$ H and $L_B = 10$ mH. The total resistance of each circuit is 40 $\Omega$. The ratio of energy consumed in circuit A and circuit B to build up the current to steady value is ______.
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If voltage and resistance are the same, the final current is the same. The energy required to "charge up" an inductor is purely dependent on its inductance size $L$!
Step 1: Understanding the Concept:
The energy stored in an inductor when the current reaches its steady-state value ($I_0$) is $U = \frac{1}{2}LI_0^2$. Step 2: Formula Application:
Steady current $I_0 = \frac{E}{R} = \frac{10}{40} = 0.25$ A.
Since $E$ and $R$ are identical for both, $I_0$ is the same for both circuits. Step 3: Explanation:
Ratio of energies $\frac{U_A}{U_B} = \frac{\frac{1}{2}L_A I_0^2}{\frac{1}{2}L_B I_0^2} = \frac{L_A}{L_B}$.
$L_A = 10$ H, $L_B = 10$ mH $= 10 \times 10^{-3}$ H.
Ratio $= \frac{10}{10 \times 10^{-3}} = 10^3 = 1000$. Step 4: Final Answer:
The ratio is 1000.