QJEEM18Ph72bb: JEE Main 2018 Physics - Class 12 JEEM18Ph72bb Solution: The equivalent capacitance between A and B in the circuit given …

Given circuit :


Simplified circuit.


The equivalent capacitance between C and D = 2 + 5 + 5 = 12 $$\mu$$F

Equivalent capacitance between E and B = 4 + 2 = 6 $$\mu$$F

Now equivalent capacitance between A and B

$$\frac{1}{C_{eq}}$$ = $$\frac{1}{6}$$ + $$\frac{1}{12}$$ + $$\frac{1}{6}$$ = $$\frac{5}{12}$$

$$\Rightarrow$$$$$$ C_eq = $$\frac{12}{5}$$ = 2.4 $$\mu$$F

Note :

(1) When capacitors C₁, C₂, . . . .C_n are in parallel then equivalent capacitance

C_eq = C₁ + C₂ + . . . . + C_n

(2) When capacitor C₁, C₂ . . . . . . C_n are in series the equivalent capacitance

$$\frac{1}{C_{eq}}$$ = $$\frac{1}{C_1}$$ + $$\frac{1}{C_2}$$ + . . . . . + $$\frac{1}{C_n}$$