Question 26: Electrostats & Capacitance

Question

Three capacitors C1, C2 and C3 are connected as shown in the figure below. If capacitor C3 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

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

Electrostats & Capacitance

Three capacitors C1, C2 and C3 are connected as shown in the figure below. If capacitor C3 breaks down electrically, the change in total charge on the combination of capacitors is

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