A resistor of 6k `Omega` with tolerance 10% and another resistance of `4kOmega` with tolerance `10%` are connected in series. The tolerance of the combination is about

Options

A. 0.05
B. 0.1
C. 0.12
D. 0.15

Explanation

To find the tolerance of the combination of two resistors connected in series, we can follow these steps:

### Step 1: Identify the values of the resistors and their tolerances
- Resistor 1 (R1) = 6 kΩ with a tolerance of 10%
- Resistor 2 (R2) = 4 kΩ with a tolerance of 10%

### Step 2: Calculate the absolute tolerances for each resistor
- The absolute tolerance (ΔR) for R1 is:
\[ \Delta R_1 = R_1 \times \left(\frac{\text{Tolerance}}{100}\right) = 6 \, \text{k}\Omega \times \left(\frac{10}{100}\right) = 0.6 \, \text{k}\Omega \]

- The absolute tolerance (ΔR) for R2 is:
\[ \Delta R_2 = R_2 \times \left(\frac{\text{Tolerance}}{100}\right) = 4 \, \text{k}\Omega \times \left(\frac{10}{100}\right) = 0.4 \, \text{k}\Omega \]

### Step 3: Calculate the total resistance of the combination
- The total resistance (R) when resistors are in series is given by:
\[ R = R_1 + R_2 = 6 \, \text{k}\Omega + 4 \, \text{k}\Omega = 10 \, \text{k}\Omega \]

### Step 4: Calculate the total tolerance for the combination
- The total tolerance (ΔR) for the combination is the sum of the absolute tolerances:
\[ \Delta R = \Delta R_1 + \Delta R_2 = 0.6 \, \text{k}\Omega + 0.4 \, \text{k}\Omega = 1 \, \text{k}\Omega \]

### Step 5: Calculate the tolerance of the combination
- The tolerance of the combination can be expressed as a percentage of the total resistance:
\[ \text{Tolerance of combination} = \frac{\Delta R}{R} \times 100 = \frac{1 \, \text{k}\Omega}{10 \, \text{k}\Omega} \times 100 = 10\% \]

### Final Answer
The tolerance of the combination is approximately **10%**.

---