Two capacitors if capacitance 0.4\(\mu F\) and 0.5\(\mu F\) are connected in parallel and charged to a p.d. of 50V, Determine the total charge acquired
Question 1 Report
Two capacitors if capacitance 0.4\(\mu F\) and 0.5\(\mu F\) are connected in parallel and charged to a p.d. of 50V, Determine the total charge acquired
Answer Details
When two capacitors are connected in parallel, their total capacitance is the sum of their individual capacitances. Therefore, the total capacitance in this case is:
C_total = C_1 + C_2 = 0.4\(\mu F\) + 0.5\(\mu F\) = 0.9\(\mu F\)
The charge acquired by each capacitor is given by:
Q = C * V
where Q is the charge, C is the capacitance, and V is the voltage.
For the 0.4\(\mu F\) capacitor, the charge acquired is:
Q_1 = C_1 * V = 0.4\(\mu F\) * 50V = 20\(\mu C\)
For the 0.5\(\mu F\) capacitor, the charge acquired is:
Q_2 = C_2 * V = 0.5\(\mu F\) * 50V = 25\(\mu C\)
Therefore, the total charge acquired is the sum of the charges acquired by each capacitor:
Q_total = Q_1 + Q_2 = 20\(\mu C\) + 25\(\mu C\) = 45\(\mu C\)
Therefore, the correct answer is 45\(\mu C\).
Download The App On Google Playstore
Everything you need to excel in your exams
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning