A wire of `15 Omega` resistance is gradually stretched to double in original length. it is then cut into two equal parts .These parts are then connected in parallel across a `3.0` volt battery. Find the current draw from the battery .

Explanation

To solve the problem step by step, we will follow the process of calculating the resistance after stretching the wire, cutting it, and then finding the current drawn from the battery.

### Step 1: Calculate the new resistance after stretching the wire
When the wire is stretched to double its original length, the resistance increases. The resistance of a wire is given by the formula:

\[ R = \rho \frac{L}{A} \]

where \( R \) is the resistance, \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. If the length is doubled, the new resistance \( R' \) can be calculated as:

\[ R' = n^2 R \]

where \( n \) is the factor by which the length is increased. Here, \( n = 2 \) and the original resistance \( R = 15 \, \Omega \).

\[ R' = 2^2 \times 15 = 4 \times 15 = 60 \, \Omega \]

### Step 2: Cut the wire into two equal parts
When the wire is cut into two equal parts, the resistance of each part \( R_{part} \) can be calculated as:

\[ R_{part} = \frac{R'}{2} = \frac{60 \, \Omega}{2} = 30 \, \Omega \]

### Step 3: Connect the two parts in parallel
Now, we connect the two 30 Ω resistors in parallel. The formula for the equivalent resistance \( R_{eq} \) of two resistors in parallel \( R_1 \) and \( R_2 \) is:

\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]

Substituting \( R_1 = 30 \, \Omega \) and \( R_2 = 30 \, \Omega \):

\[ \frac{1}{R_{eq}} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15} \]

Thus, the equivalent resistance is:

\[ R_{eq} = 15 \, \Omega \]

### Step 4: Calculate the current drawn from the battery
Now, we can find the current \( I \) drawn from the battery using Ohm's law:

\[ I = \frac{V}{R_{eq}} \]

where \( V = 3 \, V \) (the voltage of the battery) and \( R_{eq} = 15 \, \Omega \):

\[ I = \frac{3 \, V}{15 \, \Omega} = 0.2 \, A \]

### Final Answer
The current drawn from the battery is \( 0.2 \, A \).

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