Two metal wires of identical dimesnios are connected in series. If `sigma_(1)` and `sigma_(2)` are the conducties of the metal wires respectively, the effective conductivity of the combination is

Options

A. `(2sigma_(1)sigma_(2))/(sigma_(1)+sigma_(2))`
B. `(sigma_(1)+sigma_(2))/(2sigma_(1)sigma_(2))`
C. `(sigma_(1)+sigma_(2))/(sigma_(1)sigma_(2))`
D. `(sigma_(1)sigma_(2))/(sigma_(1)+sigma_(2))`

Answer

To solve the problem of finding the effective conductivity of two metal wires connected in series, we can follow these steps:

### Step 1: Understand the Setup
We have two metal wires connected in series, both having identical dimensions (length \(L\) and cross-sectional area \(A\)). The conductivities of the two wires are denoted as \(\sigma_1\) and \(\sigma_2\).

**Hint:** Remember that in a series connection, the total resistance is the sum of the individual resistances.

### Step 2: Write the Resistance Formula
The resistance \(R\) of a wire can be expressed using the formula:
\[ R = \frac{\rho L}{A} \]
where \(\rho\) is the resistivity of the material. Since conductivity \(\sigma\) is the reciprocal of resistivity, we have:
\[ \rho = \frac{1}{\sigma} \]
Thus, the resistance can be rewritten as:
\[ R = \frac{L}{\sigma A} \]

**Hint:** Make sure to express the resistance in terms of conductivity.

### Step 3: Calculate Individual Resistances
For wire 1 with conductivity \(\sigma_1\):
\[ R_1 = \frac{L}{\sigma_1 A} \]
For wire 2 with conductivity \(\sigma_2\):
\[ R_2 = \frac{L}{\sigma_2 A} \]

**Hint:** Substitute the values of \(R_1\) and \(R_2\) into the total resistance formula.

### Step 4: Find the Total Resistance
Since the wires are in series, the total resistance \(R_{total}\) is:
\[ R_{total} = R_1 + R_2 = \frac{L}{\sigma_1 A} + \frac{L}{\sigma_2 A} \]
Factoring out the common terms gives:
\[ R_{total} = \frac{L}{A} \left(\frac{1}{\sigma_1} + \frac{1}{\sigma_2}\right) \]

**Hint:** Look for a common denominator when adding fractions.

### Step 5: Express Effective Conductivity
The effective resistance \(R_{eff}\) can also be expressed in terms of effective conductivity \(\sigma_{eff}\):
\[ R_{eff} = \frac{L}{\sigma_{eff} A} \]
Setting the two expressions for resistance equal gives:
\[ \frac{L}{\sigma_{eff} A} = \frac{L}{A} \left(\frac{1}{\sigma_1} + \frac{1}{\sigma_2}\right) \]

**Hint:** Cancel out the common terms \(L\) and \(A\) from both sides.

### Step 6: Solve for Effective Conductivity
This simplifies to:
\[ \frac{1}{\sigma_{eff}} = \frac{1}{\sigma_1} + \frac{1}{\sigma_2} \]
Taking the reciprocal gives:
\[ \sigma_{eff} = \frac{2\sigma_1 \sigma_2}{\sigma_1 + \sigma_2} \]

**Hint:** This is a standard result for the effective conductivity of two conductors in series.

### Final Answer
The effective conductivity of the combination is:
\[ \sigma_{eff} = \frac{2\sigma_1 \sigma_2}{\sigma_1 + \sigma_2} \]