Question

In an experiment, two capacities measured are (13±0.1)μF and (2.4±0.2)μF. Calculate the total capacity in parallel with percentage error.​

Answer 1

⇝ Given :-

• 1st Capacity = (13 ± 0.1) μF

• 2nd Capacity = (2.4 ± 0.2) μF

⇝ To Find :-

• The total capacity in parallel with percentage error.

⇝ Solution :-

❒ For 1st Capacity :-

• C₁ = 13 μF

• Δ₁ = 0.1 μF

❒ For 2nd Capacity :-

• C₂ = 2.4 μF

• Δ₂ = 0.2 μF

We Know,

For Parralel Combination :-

$\large \underbrace{\orange{\underline{\bf C_{net}=C_1 + C_2 }}}$

Hence,

$\text C_{\text{net}}=13 + 2.4 \mu \text F$

$\purple{ \large :\longmapsto  \underline {\boxed{{\pmb{ C_{net} = 15.4 \mu\text F} }}}}$

❒ Calculating Percentage Error :

We Have,

$\Delta \text C = \Delta_ 1 + \Delta_2$

$= 0.1 + 0.2$

$\purple{ \large :\longmapsto  \underline {\boxed{{\pmb{ \Delta C = 0.3\:\mu\text F}} }}}$

Now,

We Know

$\large\underbrace{\bf\orange{\underline{\% \: error = \frac{\Delta C}{C} \times 100}}}$

Hence,

$:\longmapsto \% \: \text{error} = \frac{ 0.3}{15.4} \times 100$

$= \frac{300}{154}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\pmb{\% \: \bold{error} = 1.94\%}} }}}$

Answer 2

Refer the attachment for solution