Two inductors `L_(1)`(inductors 1 mH, internal resistance 3 `Omega`) and `L_(2)` (inductance 2mH, internal resistance 4`Omega`),and a resistor R(resistance`12 omega`) are all connected in parallelacross a 5 V battery. The circuit is switched on at time t=0. The ratio of the maximum to the minimum current `(I_(max)//I_(min))` drawn from the battery is

Explanation

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.

### Step-by-Step Solution:

1. **Identify the Components and Their Values:**
- Inductor \( L_1 \): Inductance \( = 1 \, \text{mH} \), Internal Resistance \( r_1 = 3 \, \Omega \)
- Inductor \( L_2 \): Inductance \( = 2 \, \text{mH} \), Internal Resistance \( r_2 = 4 \, \Omega \)
- Resistor \( R = 12 \, \Omega \)
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