Two capacitors marked 10μF, 400 volt and 4uF, 100 volt are connected in series. What is the maximum safe potential that can be applied when these capacitor are joined in series.
Answer & explanation
Correct answer: option 4
As the capacitors are in series so the charge on both remains same. Charge on small capacitance 4μF is Q=4×10−6×100=400μC.
Thus we can not apply charge greater than 400μC. The potential of capacitance 10μF for this charge is V=400/10=40V.
Thus the maximum potential applied is (100+40)=140V