A capacitor of 5 microfarads is connected to an A.C source of frequency 50Hz. The capacitive reactance is:
This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Convert the capacitance from microfarads to farads.
$$ C = 5 \text{ } \mu\text{F} = 5 \times 10^{-6} \text{ F} $$
Step 2: Use the formula for capacitive reactance.
$$ X_C = \frac{1}{2 \pi f C} $$
Step 3: Substitute the given values into the formula.
$$ X_C = \frac{1}{2 \pi (50 \text{ Hz}) (5 \times 10^{-6} \text{ F})} $$
$$ X_C = \frac{1}{2 \pi (250 \times 10^{-6})} $$
$$ X_C = \frac{1}{0.001570796} $$
$$ X_C \approx 636.62 \text{ } \Omega $$
The capacitive reactance is approximately 636.62 $\Omega$.
$$ \boxed{636.62 \text{ } \Omega} $$
