Four capacitors of capacitances 2 F , 3 F , 4 F and x F are connected to a battery of emf 6 ~V and of negligible internal resistance, as shown in the figure. If the ratio of the charges on x F and 4 F capacitances is 3 8 , then the value of x is

Step-by-step solution

As x and z are in parallel Let charges on 2 F , 3 F , 4 F and x F are q₂, q₃ , q₄ and q_x . q₄=C₄ V=4 F 6=24 C As, given q_x q₄ = 3 8 q_x=24 3 8 =9 C Potential difference on x F V₁= q_x x = 9 x V= q C Same 9 x volt will be across 2 F too. Now, remaining (6- 9 x ) volt potential difference will drop across 3 F . As 3 F and (x+2) F are in series, so charge on them will be same aligned q₃= & q_ (x+2) 3 (6- 9 x )= & (x+2) 9 x & 6 x-9=3 x+6 3 x=15 x=5 aligned