Question

Three capacitors $2\mu F,3\mu F$ and $5\mu F$ can withstand voltages to $3V,2V$ and $1V$ respectively. Their series combination can withstand a maximum voltage equal to :

• A. $5Volts$
• B. $\frac{31}{6}Volts$
• C. $\frac{26}{5}Volts$
• D. $None$

Answer 1

Answer: (B)

$\frac{31}{6}Volts$

R.E.F image

Solution :-

Maximum charge which flow in circuit

Shown in fig(i) will be, Q = CV

$Q=(2\times {10}^{-6}\times 3)=6\mu c$

Similarly for (ii) and (iii)

$Q_2=(3\times 2)\mu c=6\mu c$

$Q_3=(5\times 1)\mu c=5\mu c$

From above It is clear that maximum charge flow in circuit If all

the three capacitor connected in retires

should, $Q=5\mu c$

$\frac{1}{ceq}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}=\frac{15+10+6}{30}$

$=\frac{31}{30}\Rightarrow ceq=\frac{30}{31}\mu f$

$\Rightarrow Q=CV\Rightarrow 5\times {10}^{-6}=\frac{30}{31}\times {10}^{-6}\times V$

$V=\frac{31}{30}\times 5$

$V=\frac{31}{6}V$

Correct option is (B)