Question

Two wires A and B are of the same metal and of same length have their areas of cross section in the ratio 2:1 if the same potential difference is applied across each wire in turn, what will be the ratio of current flowing in A & B ?

Answer 1

The ratio of the current flowing in wire A to wire B is 2:1.

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Detailed Explanation

Let's break down the problem using the relevant physics principles.

1. List the Given Information

We are given two wires, A and B, with the following properties:
* Same Metal: This means their resistivity ($\rho$) is the same.

*   $\rho_A = \rho_B = \rho$

• Same Length:
  • $L_A = L_B = L$
• Ratio of Cross-Sectional Areas: The area of A is twice the area of B.
  • $A_A : A_B = 2:1$ or $\frac{A_A}{A_B} = \frac{2}{1}$
• Same Potential Difference (Voltage):
  • $V_A = V_B = V$

2. Identify the Goal

We need to find the ratio of the currents flowing through the wires.
* Find: $I_A : I_B$ or $\frac{I_A}{I_B}$

3. Use the Relevant Formulas

We need two key formulas:
Ohm's Law: $V = I \times R$ (Voltage = Current × Resistance)
Resistance Formula: $R = \rho \frac{L}{A}$ (Resistance = Resistivity × Length / Area)

4. Step-by-Step Calculation

Step 1: Find the relationship between current and resistance.

From Ohm's Law, we can express current as $I = \frac{V}{R}$.
Since the potential difference ($V$) is the same for both wires:
For wire A: $I_A = \frac{V}{R_A}$
For wire B: $I_B = \frac{V}{R_B}$

Now, let's find the ratio of the currents:
$$ \frac{I_A}{I_B} = \frac{V/R_A}{V/R_B} $$
The 'V' terms cancel out, leaving us with:
$$ \frac{I_A}{I_B} = \frac{R_B}{R_A} $$
This shows that the ratio of the currents is the inverse of the ratio of their resistances.

Step 2: Find the ratio of the resistances ($R_A$ and $R_B$).

Using the resistance formula, let's write the resistance for each wire:
Resistance of A: $R_A = \rho \frac{L}{A_A}$
Resistance of B: $R_B = \rho \frac{L}{A_B}$

Now, let's find the ratio of their resistances:
$$ \frac{R_A}{R_B} = \frac{\rho \frac{L}{A_A}}{\rho \frac{L}{A_B}} $$
Since $\rho$ and $L$ are the same for both wires, they cancel out:
$$ \frac{R_A}{R_B} = \frac{1/A_A}{1/A_B} = \frac{A_B}{A_A} $$
We are given that $\frac{A_A}{A_B} = \frac{2}{1}$. Therefore, the inverse ratio is:
$$ \frac{R_A}{R_B} = \frac{A_B}{A_A} = \frac{1}{2} $$

Step 3: Calculate the ratio of the currents.

From Step 1, we know that $\frac{I_A}{I_B} = \frac{R_B}{R_A}$.
From Step 2, we found that $\frac{R_A}{R_B} = \frac{1}{2}$, which means $\frac{R_B}{R_A} = \frac{2}{1}$.

Substituting this result back into our current ratio equation:
$$ \frac{I_A}{I_B} = \frac{2}{1} $$

Conclusion

The ratio of the current flowing in wire A to wire B is 2:1.

Intuitive Check: Wire A has a larger cross-sectional area (it's "thicker"). A thicker wire has less resistance to the flow of electrons. Since the same voltage (electrical "pressure") is applied to both, more current will flow through the path of lower resistance, which is wire A. Our result confirms this: the current in A is twice the current in B.