A resistor of `10kOmega` has a tolerance of 10% and another resistor of `20kOmega` has a tolerance of `20%`. The tolerance of the series combination is rearly

Options

A. `10%`
B. `13%`
C. `17%`
D. `20%`

Answer

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

### Step 1: Identify the given values
- Resistor 1 (R1) = 10 kΩ with a tolerance of 10%
- Resistor 2 (R2) = 20 kΩ with a tolerance of 20%

### Step 2: Calculate the absolute tolerances
1. **For Resistor 1 (R1)**:
- Tolerance percentage = 10%
- Absolute tolerance (ΔR1) = (Tolerance % / 100) × R1
- ΔR1 = (10 / 100) × 10 kΩ = 1 kΩ

2. **For Resistor 2 (R2)**:
- Tolerance percentage = 20%
- Absolute tolerance (ΔR2) = (Tolerance % / 100) × R2
- ΔR2 = (20 / 100) × 20 kΩ = 4 kΩ

### Step 3: Calculate the equivalent resistance in series
- The equivalent resistance (R_eq) when resistors are in series is given by:
\[ R_{eq} = R1 + R2 = 10 kΩ + 20 kΩ = 30 kΩ \]

### Step 4: Calculate the total absolute tolerance for the series combination
- The total absolute tolerance (ΔR_eq) for resistors in series is the sum of their absolute tolerances:
\[ ΔR_{eq} = ΔR1 + ΔR2 = 1 kΩ + 4 kΩ = 5 kΩ \]

### Step 5: Calculate the tolerance percentage of the equivalent resistance
- The tolerance percentage of the equivalent resistance is given by:
\[ \text{Tolerance %} = \left(\frac{ΔR_{eq}}{R_{eq}}\right) \times 100 \]
- Substituting the values:
\[ \text{Tolerance %} = \left(\frac{5 kΩ}{30 kΩ}\right) \times 100 = \frac{5}{30} \times 100 = \frac{500}{30} \approx 16.67\% \]

### Step 6: Round off the result
- Approximately, this value can be rounded to 17%.

### Conclusion
The tolerance of the series combination of the two resistors is nearly **17%**.

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