If two capacitors of 10 µF each are connected in series combination, then their equivalentcapacitance will be
Correct answer: C. 5 µF.
- A. 0 µF.
- B. 1 µF.
- C. 5 µF.
- D. 20 µF.
Explanation
When two capacitors, each with a capacitance of 10 µF, are connected in series, the formula for their equivalent capacitance is 1/C_eq = 1/C_1 + 1/C_2. Substituting the given values, we have 1/C_eq = 1/10 + 1/10 = 2/10, which simplifies to 1/C_eq = 1/5. Therefore, C_eq = 5 µF. This is less than the capacitance of each individual capacitor, which is a characteristic of capacitors in series. The other options are incorrect because they either misapply the series formula or confuse the series and parallel configurations.
Last updated
About Capacitors
Capacitors store electric charge and energy in an electric field between conductors. Work includes capacitance, the relation Q = CV, dielectric materials, charging and discharging, energy formulas, and combinations of capacitors in series and parallel. Electric potential difference is needed to understand why charge and stored energy change.
Practise Electrostatics
831 free Electrostatics MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Physics questions like this
Physics is on 13 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
A 100µF capacitor with a 12V source in series having frequency 50Hz will offer a capacitive reactance of about______________?
A 18.0 V battery is connected to a capacitor, resulting in 27.0 µC of charge stored on the capacitor. How much energy is stored in the capacitor?
A 2uF capacitor is connected to a 10 V dc supply, if 'Q' represents the charge on each plate and 'E' the energy stored on each plate, which of the following represents the values of Q and E?
A battery is permanently connected to a parallel plate capacitor and the energy stored is x joules. When one plate is moved so that the separation of the plate is doubled, the energy now stored in the joule is:
A capacitor of capacitance 30µF is charged by a constant current of 10mA. If initially, the capacitor was uncharged what is the time taken for the potential difference across the capacitor to reach 300V ?