Three capacitors connected in series have an effective capacitance of 4 μ F. If one of the capacitance is removed, the net capacitance of the capacitor increases to 6μ F.. The removed capacitor has a capacitance of

Question

Three capacitors connected in series have an effective capacitance of 4 μ F. If one of the capacitance is removed, the net capacitance of the capacitor increases to 6μ F.. The removed capacitor has a capacitance of (A) 2 μ F. (B) 4 μ F. (C) 10 μ F. (D) 12 μ F.

Tardigrade Admin · Accepted Answer

We knows when three capacitors connected in series (1/C)=(1/C1)+(1/C2)+(1/C3) Here, (1/C1)+(1/C2)+(1/C3)=(1/4) Now, on remÃ³ving one capacitÆ¡ the resultant is (1/C1)+(1/C2)=(1/6) So, (1/6)+(1/C3) =(1/4) (1/C3)=(1/4)-(1/6) =(6-4/24)=(2/24)=(1/12) C3 =12 μ F

Q. Three capacitors connected in series have an effective capacitance of 4 $μ F .$ If one of the capacitance is removed, the net capacitance of the capacitor increases to 6 $μ F .$ . The removed capacitor has a capacitance of

$2 μ F .$

$4 μ F .$

$10 μ F .$

$12 μ F .$

$24 μ F .$

Q. Three capacitors connected in series have an effective capacitance of 4 μF. If one of the capacitance is removed, the net capacitance of the capacitor increases to 6μF.. The removed capacitor has a capacitance of

2μF.

4μF.

10μF.

12μF.

24μF.