Question

If two resistors of resistances $R_{1}=(4\pm0.5)\Omega$ and $R_{2}=(16\pm0.5)\Omega$ are connected in series. The equivalent resistance with the limits of percentage error is ________

• A. $(20\pm1%)\Omega$
• B. $(20\pm5%)\Omega$
• C. $(20\pm0.25%)\Omega$
• D. $(20\pm0.5%)\Omega$

Hint:

Errors always add up in addition or subtraction.

Answer 1

The Correct Option is B

Step 1: Concept
In series, $R_{eq} = R_1 + R_2$ and absolute errors add up: $\Delta R_{eq} = \Delta R_1 + \Delta R_2$.
Step 2: Analysis
- $R_{eq} = 4 + 16 = 20\Omega$. - $\Delta R_{eq} = 0.5 + 0.5 = 1.0\Omega$.
Step 3: Calculation
- Percentage Error = $(\frac{\Delta R_{eq}}{R_{eq}}) \times 100%$. - Percentage Error = $(\frac{1.0}{20}) \times 100% = 5%$.
Step 4: Conclusion
Hence, the equivalent resistance is $(20 \pm 5%)\Omega$.
Final Answer:(B)