To solve the problem, we need to find the ratio of the maximum to the minimum current drawn from the battery when two inductors and a resistor are connected in parallel. Let's go through the solution step by step.
### Step 1: Identify the Components
We have:
- Inductor \( L_1 \) with inductance \( 1 \, \text{mH} \) and internal resistance \( 3 \, \Omega \)
- Inductor \( L_2 \) with inductance \( 2 \, \text{mH} \) and internal resistance \( 4 \, \Omega \)
- Resistor \( R \) with resistance \( 12 \, \Omega \)
- Voltage of the battery \( E = 5 \, \text{V} \)
### Step 2: Determine the Equivalent Resistance at \( t = 0 \)
At time \( t = 0 \), the inductors behave like open circuits because they have not yet built up any current. Therefore, the only resistance in the circuit is the resistor \( R \).
\[
R_{\text{eq, max}} = R = 12 \, \Omega
\]
### Step 3: Determine the Equivalent Resistance as \( t \to \infty \)
As time approaches infinity, the inductors behave like short circuits (i.e., they have zero resistance). We need to find the equivalent resistance of the circuit when both inductors are shorted.
The equivalent resistance \( R_{\text{eq, min}} \) for the parallel combination of the resistances is given by:
\[
\frac{1}{R_{\text{eq, min}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R}
\]
Where:
- \( R_1 = 3 \, \Omega \) (internal resistance of \( L_1 \))
- \( R_2 = 4 \, \Omega \) (internal resistance of \( L_2 \))
- \( R = 12 \, \Omega \) (external resistance)
Substituting the values:
\[
\frac{1}{R_{\text{eq, min}}} = \frac{1}{3} + \frac{1}{4} + \frac{1}{12}
\]
Finding a common denominator (which is 12):
\[
\frac{1}{R_{\text{eq, min}}} = \frac{4}{12} + \frac{3}{12} + \frac{1}{12} = \frac{8}{12}
\]
Thus,
\[
R_{\text{eq, min}} = \frac{12}{8} = 1.5 \, \Omega
\]
### Step 4: Calculate Maximum and Minimum Currents
Using Ohm's Law, the current drawn from the battery can be calculated as:
- Maximum current \( I_{\text{max}} \) when \( R_{\text{eq}} = R_{\text{eq, max}} \):
\[
I_{\text{max}} = \frac{E}{R_{\text{eq, max}}} = \frac{5 \, \text{V}}{12 \, \Omega} = \frac{5}{12} \, \text{A}
\]
- Minimum current \( I_{\text{min}} \) when \( R_{\text{eq}} = R_{\text{eq, min}} \):
\[
I_{\text{min}} = \frac{E}{R_{\text{eq, min}}} = \frac{5 \, \text{V}}{1.5 \, \Omega} = \frac{5}{1.5} = \frac{10}{3} \, \text{A}
\]
### Step 5: Calculate the Ratio of Maximum to Minimum Current
Now, we can find the ratio of maximum to minimum current:
\[
\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{\frac{5}{12}}{\frac{10}{3}} = \frac{5}{12} \times \frac{3}{10} = \frac{15}{120} = \frac{1}{8}
\]
Thus, the ratio of maximum to minimum current is:
\[
\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{1}{8}
\]
### Final Answer
The ratio of the maximum to the minimum current drawn from the battery is \( \frac{1}{8} \).
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