If two wires P and O each of the same length and same material,are connected in parallel to a battery. The diameter of P is half that of Q. What fraction of the total current passes through P?(a)0.2(b)0.25(c)0.33(d)0.5

The same current flows through the same cross-sectional area, but with only half the diameter, P has only ____ the area. Thus it passes ___ the current.

[If you halve the diameter what happens to the area?]

The answer is 0.20

use formula A = (pi*(d^2))/4
As resistivity and length are equal, they will cancel each other The ratio of resistance P to resistance Q will be 4.0.
I_p = V/R_p I_Q = V/R_Q
I_total = I_p + I_Q
dividing I_p by I , you will get 1/5 which is equal to 0.20

To determine the fraction of the total current that passes through wire P, we need to consider the relationship between the resistance and the current in parallel circuits.

1. In parallel circuits, the total current is divided among the different branches based on the resistance of each branch. The branch with less resistance will have a larger fraction of the total current passing through it.

2. The resistance of a wire can be calculated using the formula: R = ρ * (L/A), where R is the resistance, ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.

3. Since wires P and O have the same length and material, their resistances will depend only on their cross-sectional areas.

4. The cross-sectional area of wire P is half that of wire O, which means the resistance of wire P will be twice that of wire O.

5. Therefore, wire O will have half the resistance of wire P.

6. Based on Ohm's Law (V = I * R), the current passing through a wire is inversely proportional to its resistance. So, wire O will carry twice the current as wire P.

7. Let's assume the total current passing through the wires is I_t.

8. Since wire O carries twice the current as wire P, the current passing through wire O will be 2/3 * I_t and the current passing through wire P will be 1/3 * I_t.

9. Therefore, the fraction of the total current passing through wire P is 1/3, which is equivalent to 0.33.

Therefore, the correct answer is (c) 0.33.

To determine the fraction of the total current that passes through wire P, we can make use of the formula:

\(I_P = (R_P / R_T ) × I_T \)

Where:
- \(I_P\) is the current passing through wire P,
- \(R_P\) is the resistance of wire P,
- \(R_T\) is the total resistance of both wires connected in parallel,
- \(I_T\) is the total current passing through the parallel combination.

Since wires P and O are made of the same material and have the same length, the resistance of each wire is directly proportional to its length and inversely proportional to the square of its diameter.

Let's assume:
- The length of wires P and O is L,
- The diameter of wire P is D,
- The diameter of wire O is 2D (since the diameter of P is half that of Q).

Given that wires P and O are connected in parallel to a battery, the total resistance is given by:

\(1/R_T = 1/R_P + 1/R_O \)

Substituting the resistance formula into the above equation, we get:

\(1/R_T = L / (π × D^2) + L / (π × (2D)^2) \)

Simplifying this equation, we have:

\(1/R_T = L / (π × D^2) + L / (4π × D^2) \)

\(1/R_T = 5L / (4π × D^2) \)

Now, to find \(I_P/I_T\), we substitute the values into our initial formula:

\(I_P = (R_P / R_T ) × I_T \)

\(I_P = (L / (π × D^2)) / (5L / (4π × D^2)) × I_T \)

\(I_P = 4 / 5 × I_T \)

So, the fraction of the total current that passes through wire P is 4/5.

However, none of the given options are equal to 4/5. It seems there might be a mistake with the provided answer choices. If you could double-check, I'll be happy to assist you further.