In the given network capacitance C2=10 μ F, C1=5 μ F and C3=4 μ F. The resultant capacitance between P and Q will be :

Question

In the given network capacitance C2=10 μ F, C1=5 μ F and C3=4 μ F. The resultant capacitance between P and Q will be : (A)  4.7 μ F  (B)  1.2 μ F  (C)  3.2 μ F  (D)  2.2 μ F

Tardigrade Admin · Accepted Answer

Seeing in the given circuit C1 and C2 are connected in parallel. Hence, their equivalent capacitance Ce q =C1+C2=5+10 =15 μ F As Ceq and C3 are connected in series, Hence, resultant capacitance between P and Q is given by (1/CP Q)=(1/Ce q)+(1/C3) =(1/15)+(1/4)=(19/60) CP Q=(60/19)=3.2 μ F

Q. In the given network capacitance $C_{2} = 10 μ F, C_{1} = 5 μ F$ and $C_{3} = 4 μ F$ . The resultant capacitance between $P$ and $Q$ will be :

$4.7 μ F$

$1.2 μ F$

$3.2 μ F$

$2.2 μ F$

Q. In the given network capacitance C2​=10μF,C1​=5μF and C3​=4μF. The resultant capacitance between P and Q will be :

4.7μF

1.2μF

3.2μF

2.2μF

Q. In the given network capacitance $C_{2} = 10 μ F, C_{1} = 5 μ F$ and $C_{3} = 4 μ F$ . The resultant capacitance between $P$ and $Q$ will be :

$4.7 μ F$

$1.2 μ F$

$3.2 μ F$

$2.2 μ F$