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**A supply of 400V is connected across capacitors of 3µf and 6µf in series. Calculate the charge**

**QUICK ANSWER…**

**A**

**DETAILS…**

**The appropriate formula relating voltage, capacitance, and charge is given by:**

**Q = CV**

**Where Q = charge = ?**

**V = voltage across the terminals = 400V**

**C = capacitance of the capacitors which can be calculated as a resultant of the two series connected capacitors.**

**For capacitors in series,**

**Now for the right answer to the above question:**

**Option A is correct.****Option B is incorrect.****C is incorrect.****D is not the correct answer.**

**HELPFUL TIPS…**

**You may please note these/this:**

**The formula relating capacitance, charge, and voltage, is given by: Q = CV.**

**Use the questions and answer section to correct, express, and contribute your views.**

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**/culled from 2019 JAMB-UTME physics question 24/**