Question 30: Electrostatics & Capacitance

Question

Three capacitors C
1
, C
2
and C
3
are connected as shown in the figure below. If capacitor C
3
breaks down electrically, the change in total charge on the combination of capacitors is

Accepted Answer

Since capacitors C
1
and C
2
are parallel, so their equivalent capacitance will be (C
1
+ C
2
). Now, (C
1
+ C
2
) and C
3
are in series, so the net equivalent capacitance of circuit will be,
$${1 \over C} = {1 \over {{C_3}}} + {1 \over {{C_1} + {C_2}}} = {{{C_1} + {C_2} + {C_3}} \over {({C_1} + {C_2}){C_3}}}$$
or, $$C = {{({C_1} + {C_2}){C_3}} \over {{C_1} + {C_2} + {C_3}}}$$
Since V is the voltage of the battery, so charge on this system,
$$q = CV \Rightarrow q = {{({C_1} + {C_2}){C_3}V} \over {({C_1} + {C_2} + {C_3})}}$$
If the capacitance C
3
breaks down, then total equivalent capacitance will become
$$C' = {C_1} + {C_2}$$
$$	herefore$$ New charge stored,
$$q' = C'V$$
$$ \Rightarrow q' = ({C_1} + {C_2})V$$
Now, change in total charge,
$$\Delta q = q' - q$$ [$$\because$$ $$q' > q$$]
$$ = ({C_1} + {C_2})V - {{({C_1} + {C_2}){C_3}V} \over {{C_1} + {C_2} + {C_2}}}$$
or, $$\Delta q = ({C_1} + {C_2})V\left[ {1 - \left( {{{{C_3}} \over {{C_1} + {C_2} + {C_3}}}} ight)} ight]$$

Answer

$$({C_1} + {C_2})V\left[ {1 - \left( {{{{C_3}} \over {{C_1} + {C_2} + {C_3}}}} \right)} \right]$$

Answer

$$({C_3} + {C_2})V\left[ {1 - \left( {{{{C_1}} \over {{C_1} + {C_2} + {C_3}}}} \right)} \right]$$

Answer

$$({C_1} + {C_2})V\left[ {1 - \left( {{{{C_2} + {C_1}} \over {{C_1} + {C_3}}}} \right)} \right]$$

Answer

$$({C_1} + {C_2})V\left[ {1 - \left( {{{{C_2}} \over {{C_1} + {C_3}}}} \right)} \right]$$

Electrostatics & Capacitance

Three capacitors C 1 , C 2 and C 3 are connected as shown in the figure below. If capacitor C 3 breaks down electrically, the change in total charge on the combination of capacitors is

Choose the correct answer: