If each capacitor of 18 μ F arranged as shown in figure below, then the equivalent capacitance between the points P and Q

Question

If each capacitor of 18 μ F arranged as shown in figure below, then the equivalent capacitance between the points P and Q <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/714e70f9489b017d04bdf770e5e858dc-.png" /> (A) 30 μ F (B) 90 μ F (C) 20 μ F (D) 18 μ F

Tardigrade Admin · Accepted Answer

CP=C1+C2=18+18=36 μ F <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e33eeb405656d32904e23c39c363c6f5-.png" /> (1/Cs)=(1/36)+(1/18)=(1+2/36) Cs=12 μ F CP=12+18=30 μ F

Q. If each capacitor of $18 μ F$ arranged as shown in figure below, then the equivalent capacitance between the points $P$ and $Q$

$30 μ F$

$90 μ F$

$20 μ F$

$18 μ F$

$C_{P} = C_{1} + C_{2} = 18 + 18 = 36 μ F$

$\frac{1}{C _{s}} = \frac{1}{36} + \frac{1}{18} = \frac{1 + 2}{36}$
$C_{s} = 12 μ F$
$C_{P} = 12 + 18 = 30 μ F$

Q. If each capacitor of 18μF arranged as shown in figure below, then the equivalent capacitance between the points P and Q

30μF

90μF

20μF

18μF