NEET Test Series from KOTA - 10 Papers In MS WORD
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Capacitance
165833
Five equal capacitors each with capacitance $C$ are connected as shown in figure. Then, the equivalent capacitance between $A$ and $B$ is
1 $5 \mathrm{C}$
2 $\frac{\mathrm{C}}{5}$
3 $3 \mathrm{C}$
4 $\mathrm{C}$
Explanation:
: Given figure, We can remove the capacitor between C and D It is balanced Wheatstone bridge. $\mathrm{C}_{\mathrm{AB}} =\frac{\mathrm{C}}{2}+\frac{\mathrm{C}}{2}$ $\mathrm{C}_{\mathrm{AB}} =\mathrm{C}$
[Manipal UGET -2020
Capacitance
165834
Effective capacitance between $A$ and $B$ in the figure shown is (all capacitances are in $\mu \mathrm{F}$ )
1 $\frac{3}{14} \mu \mathrm{F}$
2 $\frac{14}{3} \mu \mathrm{F}$
3 $21 \mu \mathrm{F}$
4 $23 \mu \mathrm{F}$
Explanation:
: Given figure, Balance Wheatstone bridge- $\frac{3}{4}=\frac{6}{8}$ The point $\mathrm{C}$ and $\mathrm{D}$ has no charge- $\mathrm{C}_{1}=\frac{3 \times 6}{3+6}=\frac{18}{9}=2 \mu \mathrm{F}$ and $\mathrm{C}_{2}=\frac{4 \times 8}{4+8}=\frac{32}{12}=\frac{8}{3} \mu \mathrm{F}$ $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+\frac{8}{3}=\frac{6+8}{3}=\frac{14}{3} \mu \mathrm{F}$
[Manipal UGET-2012]
Capacitance
165835
The charge deposited on $4 \mu \mathrm{F}$ capacitor in the circuit is
1 $6 \times 10^{-6} \mathrm{C}$
2 $12 \times 10^{-6} \mathrm{C}$
3 $24 \times 10^{-6} \mathrm{C}$
4 $36 \times 10^{-6} \mathrm{C}$
Explanation:
: Given that, As the capacitor $4 \mu \mathrm{F} \& 2 \mu \mathrm{F}$ connected in parallel then its resultant are in series with $6 \mu \mathrm{F}$ capacitor, $\mathrm{C}_{\text {eq }}=\frac{(2+4) \times 6}{2+4+6}=3 \mu \mathrm{F}$ Charge in circuit- $\mathrm{Q}=3 \mu \mathrm{F} \times 12 \mathrm{~V}=36 \mu \mathrm{C}$ $\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{4}{2}$ $\mathrm{Q}_{1}=2 \mathrm{Q}_{2}$ Hence, $\mathrm{Q}=\mathrm{Q}_{1}+\mathrm{Q}_{2}$ $36 \mu \mathrm{C}=2 \mathrm{Q}_{2}+\mathrm{Q}_{2}$ $\mathrm{Q}_{2}=\frac{36 \mu \mathrm{C}}{3}=12 \mu \mathrm{F}$ Hence, $\mathrm{Q}_{1}=\mathrm{Q}-\mathrm{Q}_{2}$ $\mathrm{Q}_{1}=36 \mu \mathrm{C}-12 \mu \mathrm{C}$ $\mathrm{Q}_{1}=24 \times 10^{-6} \mathrm{C}$
[Manipal UGET-2009]
Capacitance
165836
In the following circuit the resultant capacitance between $A$ and $B$ is $1 \mu \mathrm{F}$. Then, the value of $C$ is
1 $\frac{32}{11} \mu \mathrm{F}$
2 $\frac{11}{32} \mu \mathrm{F}$
3 $\frac{23}{32} \mu \mathrm{F}$
4 $\frac{32}{23} \mu \mathrm{F}$
Explanation:
: Given, Capacitance between $\mathrm{A}$ and $\mathrm{B}, \mathrm{C}_{\mathrm{AB}}=1 \mu \mathrm{F}$. $\therefore$ Capacitors between point $\mathrm{C}$ and $\mathrm{D}$ are connected in parallel so capacitance. $\mathrm{C}_{\mathrm{CD}}=2 \mu \mathrm{F}+2 \mu \mathrm{F}=4 \mu \mathrm{F}$ $\therefore$ Capacitors between point $\mathrm{E}$ and $\mathrm{F}$ connected in series, so the capacitance between points $\mathrm{E}$ and $\mathrm{F}$, is $\frac{1}{\mathrm{C}_{\mathrm{EF}}}=\frac{1}{6}+\frac{1}{12}=\frac{2+1}{12}=\frac{3}{12}$ $\mathrm{C}_{\mathrm{EF}}=\frac{12}{3}=4 \mu \mathrm{F}$ Capacitors between points $G$ and $D$ are connected in series, therefore, $\frac{1}{\mathrm{C}_{\mathrm{GD}}}=\frac{1}{8}+\frac{1}{4}=\frac{1+2}{8}=\frac{3}{8}$ $\mathrm{C}_{\mathrm{GD}}=\frac{8}{3} \mu \mathrm{F}$ $\therefore$ Capacitance between points $\mathrm{E}$ and $\mathrm{F}$, are connected in parallel, So Capacitors between points $\mathrm{G}$ and $\mathrm{F}$, are connected in series, therefore, $\frac{1}{\mathrm{C}_{\mathrm{GF}}}=\frac{1}{1}+\frac{1}{8}=\frac{8+1}{8}=\frac{9}{8}$ $\mathrm{C}_{\mathrm{GF}}=\frac{8}{9} \mu \mathrm{F}$ $\mathrm{C}_{\mathrm{GD}}=\frac{8}{3}+\frac{8}{9}=\frac{24+8}{9}=\frac{32}{9} \mu \mathrm{F}$ $\frac{1}{\mathrm{C}_{\mathrm{AB}}}=\frac{1}{\mathrm{C}}+\frac{1}{32 / 9}$ $\frac{1}{\mathrm{C}_{\mathrm{AB}}}=\frac{1}{\mathrm{C}}+\frac{9}{32}$ Here, $\mathrm{C}_{\mathrm{AB}}=1 \mu \mathrm{F}$ $\frac{1}{\mathrm{C}}=1-\frac{9}{32}=\frac{32-9}{32}=\frac{23}{32}$ $\mathrm{C}=\frac{32}{23} \mu \mathrm{F}$
165833
Five equal capacitors each with capacitance $C$ are connected as shown in figure. Then, the equivalent capacitance between $A$ and $B$ is
1 $5 \mathrm{C}$
2 $\frac{\mathrm{C}}{5}$
3 $3 \mathrm{C}$
4 $\mathrm{C}$
Explanation:
: Given figure, We can remove the capacitor between C and D It is balanced Wheatstone bridge. $\mathrm{C}_{\mathrm{AB}} =\frac{\mathrm{C}}{2}+\frac{\mathrm{C}}{2}$ $\mathrm{C}_{\mathrm{AB}} =\mathrm{C}$
[Manipal UGET -2020
Capacitance
165834
Effective capacitance between $A$ and $B$ in the figure shown is (all capacitances are in $\mu \mathrm{F}$ )
1 $\frac{3}{14} \mu \mathrm{F}$
2 $\frac{14}{3} \mu \mathrm{F}$
3 $21 \mu \mathrm{F}$
4 $23 \mu \mathrm{F}$
Explanation:
: Given figure, Balance Wheatstone bridge- $\frac{3}{4}=\frac{6}{8}$ The point $\mathrm{C}$ and $\mathrm{D}$ has no charge- $\mathrm{C}_{1}=\frac{3 \times 6}{3+6}=\frac{18}{9}=2 \mu \mathrm{F}$ and $\mathrm{C}_{2}=\frac{4 \times 8}{4+8}=\frac{32}{12}=\frac{8}{3} \mu \mathrm{F}$ $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+\frac{8}{3}=\frac{6+8}{3}=\frac{14}{3} \mu \mathrm{F}$
[Manipal UGET-2012]
Capacitance
165835
The charge deposited on $4 \mu \mathrm{F}$ capacitor in the circuit is
1 $6 \times 10^{-6} \mathrm{C}$
2 $12 \times 10^{-6} \mathrm{C}$
3 $24 \times 10^{-6} \mathrm{C}$
4 $36 \times 10^{-6} \mathrm{C}$
Explanation:
: Given that, As the capacitor $4 \mu \mathrm{F} \& 2 \mu \mathrm{F}$ connected in parallel then its resultant are in series with $6 \mu \mathrm{F}$ capacitor, $\mathrm{C}_{\text {eq }}=\frac{(2+4) \times 6}{2+4+6}=3 \mu \mathrm{F}$ Charge in circuit- $\mathrm{Q}=3 \mu \mathrm{F} \times 12 \mathrm{~V}=36 \mu \mathrm{C}$ $\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{4}{2}$ $\mathrm{Q}_{1}=2 \mathrm{Q}_{2}$ Hence, $\mathrm{Q}=\mathrm{Q}_{1}+\mathrm{Q}_{2}$ $36 \mu \mathrm{C}=2 \mathrm{Q}_{2}+\mathrm{Q}_{2}$ $\mathrm{Q}_{2}=\frac{36 \mu \mathrm{C}}{3}=12 \mu \mathrm{F}$ Hence, $\mathrm{Q}_{1}=\mathrm{Q}-\mathrm{Q}_{2}$ $\mathrm{Q}_{1}=36 \mu \mathrm{C}-12 \mu \mathrm{C}$ $\mathrm{Q}_{1}=24 \times 10^{-6} \mathrm{C}$
[Manipal UGET-2009]
Capacitance
165836
In the following circuit the resultant capacitance between $A$ and $B$ is $1 \mu \mathrm{F}$. Then, the value of $C$ is
1 $\frac{32}{11} \mu \mathrm{F}$
2 $\frac{11}{32} \mu \mathrm{F}$
3 $\frac{23}{32} \mu \mathrm{F}$
4 $\frac{32}{23} \mu \mathrm{F}$
Explanation:
: Given, Capacitance between $\mathrm{A}$ and $\mathrm{B}, \mathrm{C}_{\mathrm{AB}}=1 \mu \mathrm{F}$. $\therefore$ Capacitors between point $\mathrm{C}$ and $\mathrm{D}$ are connected in parallel so capacitance. $\mathrm{C}_{\mathrm{CD}}=2 \mu \mathrm{F}+2 \mu \mathrm{F}=4 \mu \mathrm{F}$ $\therefore$ Capacitors between point $\mathrm{E}$ and $\mathrm{F}$ connected in series, so the capacitance between points $\mathrm{E}$ and $\mathrm{F}$, is $\frac{1}{\mathrm{C}_{\mathrm{EF}}}=\frac{1}{6}+\frac{1}{12}=\frac{2+1}{12}=\frac{3}{12}$ $\mathrm{C}_{\mathrm{EF}}=\frac{12}{3}=4 \mu \mathrm{F}$ Capacitors between points $G$ and $D$ are connected in series, therefore, $\frac{1}{\mathrm{C}_{\mathrm{GD}}}=\frac{1}{8}+\frac{1}{4}=\frac{1+2}{8}=\frac{3}{8}$ $\mathrm{C}_{\mathrm{GD}}=\frac{8}{3} \mu \mathrm{F}$ $\therefore$ Capacitance between points $\mathrm{E}$ and $\mathrm{F}$, are connected in parallel, So Capacitors between points $\mathrm{G}$ and $\mathrm{F}$, are connected in series, therefore, $\frac{1}{\mathrm{C}_{\mathrm{GF}}}=\frac{1}{1}+\frac{1}{8}=\frac{8+1}{8}=\frac{9}{8}$ $\mathrm{C}_{\mathrm{GF}}=\frac{8}{9} \mu \mathrm{F}$ $\mathrm{C}_{\mathrm{GD}}=\frac{8}{3}+\frac{8}{9}=\frac{24+8}{9}=\frac{32}{9} \mu \mathrm{F}$ $\frac{1}{\mathrm{C}_{\mathrm{AB}}}=\frac{1}{\mathrm{C}}+\frac{1}{32 / 9}$ $\frac{1}{\mathrm{C}_{\mathrm{AB}}}=\frac{1}{\mathrm{C}}+\frac{9}{32}$ Here, $\mathrm{C}_{\mathrm{AB}}=1 \mu \mathrm{F}$ $\frac{1}{\mathrm{C}}=1-\frac{9}{32}=\frac{32-9}{32}=\frac{23}{32}$ $\mathrm{C}=\frac{32}{23} \mu \mathrm{F}$
165833
Five equal capacitors each with capacitance $C$ are connected as shown in figure. Then, the equivalent capacitance between $A$ and $B$ is
1 $5 \mathrm{C}$
2 $\frac{\mathrm{C}}{5}$
3 $3 \mathrm{C}$
4 $\mathrm{C}$
Explanation:
: Given figure, We can remove the capacitor between C and D It is balanced Wheatstone bridge. $\mathrm{C}_{\mathrm{AB}} =\frac{\mathrm{C}}{2}+\frac{\mathrm{C}}{2}$ $\mathrm{C}_{\mathrm{AB}} =\mathrm{C}$
[Manipal UGET -2020
Capacitance
165834
Effective capacitance between $A$ and $B$ in the figure shown is (all capacitances are in $\mu \mathrm{F}$ )
1 $\frac{3}{14} \mu \mathrm{F}$
2 $\frac{14}{3} \mu \mathrm{F}$
3 $21 \mu \mathrm{F}$
4 $23 \mu \mathrm{F}$
Explanation:
: Given figure, Balance Wheatstone bridge- $\frac{3}{4}=\frac{6}{8}$ The point $\mathrm{C}$ and $\mathrm{D}$ has no charge- $\mathrm{C}_{1}=\frac{3 \times 6}{3+6}=\frac{18}{9}=2 \mu \mathrm{F}$ and $\mathrm{C}_{2}=\frac{4 \times 8}{4+8}=\frac{32}{12}=\frac{8}{3} \mu \mathrm{F}$ $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+\frac{8}{3}=\frac{6+8}{3}=\frac{14}{3} \mu \mathrm{F}$
[Manipal UGET-2012]
Capacitance
165835
The charge deposited on $4 \mu \mathrm{F}$ capacitor in the circuit is
1 $6 \times 10^{-6} \mathrm{C}$
2 $12 \times 10^{-6} \mathrm{C}$
3 $24 \times 10^{-6} \mathrm{C}$
4 $36 \times 10^{-6} \mathrm{C}$
Explanation:
: Given that, As the capacitor $4 \mu \mathrm{F} \& 2 \mu \mathrm{F}$ connected in parallel then its resultant are in series with $6 \mu \mathrm{F}$ capacitor, $\mathrm{C}_{\text {eq }}=\frac{(2+4) \times 6}{2+4+6}=3 \mu \mathrm{F}$ Charge in circuit- $\mathrm{Q}=3 \mu \mathrm{F} \times 12 \mathrm{~V}=36 \mu \mathrm{C}$ $\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{4}{2}$ $\mathrm{Q}_{1}=2 \mathrm{Q}_{2}$ Hence, $\mathrm{Q}=\mathrm{Q}_{1}+\mathrm{Q}_{2}$ $36 \mu \mathrm{C}=2 \mathrm{Q}_{2}+\mathrm{Q}_{2}$ $\mathrm{Q}_{2}=\frac{36 \mu \mathrm{C}}{3}=12 \mu \mathrm{F}$ Hence, $\mathrm{Q}_{1}=\mathrm{Q}-\mathrm{Q}_{2}$ $\mathrm{Q}_{1}=36 \mu \mathrm{C}-12 \mu \mathrm{C}$ $\mathrm{Q}_{1}=24 \times 10^{-6} \mathrm{C}$
[Manipal UGET-2009]
Capacitance
165836
In the following circuit the resultant capacitance between $A$ and $B$ is $1 \mu \mathrm{F}$. Then, the value of $C$ is
1 $\frac{32}{11} \mu \mathrm{F}$
2 $\frac{11}{32} \mu \mathrm{F}$
3 $\frac{23}{32} \mu \mathrm{F}$
4 $\frac{32}{23} \mu \mathrm{F}$
Explanation:
: Given, Capacitance between $\mathrm{A}$ and $\mathrm{B}, \mathrm{C}_{\mathrm{AB}}=1 \mu \mathrm{F}$. $\therefore$ Capacitors between point $\mathrm{C}$ and $\mathrm{D}$ are connected in parallel so capacitance. $\mathrm{C}_{\mathrm{CD}}=2 \mu \mathrm{F}+2 \mu \mathrm{F}=4 \mu \mathrm{F}$ $\therefore$ Capacitors between point $\mathrm{E}$ and $\mathrm{F}$ connected in series, so the capacitance between points $\mathrm{E}$ and $\mathrm{F}$, is $\frac{1}{\mathrm{C}_{\mathrm{EF}}}=\frac{1}{6}+\frac{1}{12}=\frac{2+1}{12}=\frac{3}{12}$ $\mathrm{C}_{\mathrm{EF}}=\frac{12}{3}=4 \mu \mathrm{F}$ Capacitors between points $G$ and $D$ are connected in series, therefore, $\frac{1}{\mathrm{C}_{\mathrm{GD}}}=\frac{1}{8}+\frac{1}{4}=\frac{1+2}{8}=\frac{3}{8}$ $\mathrm{C}_{\mathrm{GD}}=\frac{8}{3} \mu \mathrm{F}$ $\therefore$ Capacitance between points $\mathrm{E}$ and $\mathrm{F}$, are connected in parallel, So Capacitors between points $\mathrm{G}$ and $\mathrm{F}$, are connected in series, therefore, $\frac{1}{\mathrm{C}_{\mathrm{GF}}}=\frac{1}{1}+\frac{1}{8}=\frac{8+1}{8}=\frac{9}{8}$ $\mathrm{C}_{\mathrm{GF}}=\frac{8}{9} \mu \mathrm{F}$ $\mathrm{C}_{\mathrm{GD}}=\frac{8}{3}+\frac{8}{9}=\frac{24+8}{9}=\frac{32}{9} \mu \mathrm{F}$ $\frac{1}{\mathrm{C}_{\mathrm{AB}}}=\frac{1}{\mathrm{C}}+\frac{1}{32 / 9}$ $\frac{1}{\mathrm{C}_{\mathrm{AB}}}=\frac{1}{\mathrm{C}}+\frac{9}{32}$ Here, $\mathrm{C}_{\mathrm{AB}}=1 \mu \mathrm{F}$ $\frac{1}{\mathrm{C}}=1-\frac{9}{32}=\frac{32-9}{32}=\frac{23}{32}$ $\mathrm{C}=\frac{32}{23} \mu \mathrm{F}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Capacitance
165833
Five equal capacitors each with capacitance $C$ are connected as shown in figure. Then, the equivalent capacitance between $A$ and $B$ is
1 $5 \mathrm{C}$
2 $\frac{\mathrm{C}}{5}$
3 $3 \mathrm{C}$
4 $\mathrm{C}$
Explanation:
: Given figure, We can remove the capacitor between C and D It is balanced Wheatstone bridge. $\mathrm{C}_{\mathrm{AB}} =\frac{\mathrm{C}}{2}+\frac{\mathrm{C}}{2}$ $\mathrm{C}_{\mathrm{AB}} =\mathrm{C}$
[Manipal UGET -2020
Capacitance
165834
Effective capacitance between $A$ and $B$ in the figure shown is (all capacitances are in $\mu \mathrm{F}$ )
1 $\frac{3}{14} \mu \mathrm{F}$
2 $\frac{14}{3} \mu \mathrm{F}$
3 $21 \mu \mathrm{F}$
4 $23 \mu \mathrm{F}$
Explanation:
: Given figure, Balance Wheatstone bridge- $\frac{3}{4}=\frac{6}{8}$ The point $\mathrm{C}$ and $\mathrm{D}$ has no charge- $\mathrm{C}_{1}=\frac{3 \times 6}{3+6}=\frac{18}{9}=2 \mu \mathrm{F}$ and $\mathrm{C}_{2}=\frac{4 \times 8}{4+8}=\frac{32}{12}=\frac{8}{3} \mu \mathrm{F}$ $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+\frac{8}{3}=\frac{6+8}{3}=\frac{14}{3} \mu \mathrm{F}$
[Manipal UGET-2012]
Capacitance
165835
The charge deposited on $4 \mu \mathrm{F}$ capacitor in the circuit is
1 $6 \times 10^{-6} \mathrm{C}$
2 $12 \times 10^{-6} \mathrm{C}$
3 $24 \times 10^{-6} \mathrm{C}$
4 $36 \times 10^{-6} \mathrm{C}$
Explanation:
: Given that, As the capacitor $4 \mu \mathrm{F} \& 2 \mu \mathrm{F}$ connected in parallel then its resultant are in series with $6 \mu \mathrm{F}$ capacitor, $\mathrm{C}_{\text {eq }}=\frac{(2+4) \times 6}{2+4+6}=3 \mu \mathrm{F}$ Charge in circuit- $\mathrm{Q}=3 \mu \mathrm{F} \times 12 \mathrm{~V}=36 \mu \mathrm{C}$ $\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{4}{2}$ $\mathrm{Q}_{1}=2 \mathrm{Q}_{2}$ Hence, $\mathrm{Q}=\mathrm{Q}_{1}+\mathrm{Q}_{2}$ $36 \mu \mathrm{C}=2 \mathrm{Q}_{2}+\mathrm{Q}_{2}$ $\mathrm{Q}_{2}=\frac{36 \mu \mathrm{C}}{3}=12 \mu \mathrm{F}$ Hence, $\mathrm{Q}_{1}=\mathrm{Q}-\mathrm{Q}_{2}$ $\mathrm{Q}_{1}=36 \mu \mathrm{C}-12 \mu \mathrm{C}$ $\mathrm{Q}_{1}=24 \times 10^{-6} \mathrm{C}$
[Manipal UGET-2009]
Capacitance
165836
In the following circuit the resultant capacitance between $A$ and $B$ is $1 \mu \mathrm{F}$. Then, the value of $C$ is
1 $\frac{32}{11} \mu \mathrm{F}$
2 $\frac{11}{32} \mu \mathrm{F}$
3 $\frac{23}{32} \mu \mathrm{F}$
4 $\frac{32}{23} \mu \mathrm{F}$
Explanation:
: Given, Capacitance between $\mathrm{A}$ and $\mathrm{B}, \mathrm{C}_{\mathrm{AB}}=1 \mu \mathrm{F}$. $\therefore$ Capacitors between point $\mathrm{C}$ and $\mathrm{D}$ are connected in parallel so capacitance. $\mathrm{C}_{\mathrm{CD}}=2 \mu \mathrm{F}+2 \mu \mathrm{F}=4 \mu \mathrm{F}$ $\therefore$ Capacitors between point $\mathrm{E}$ and $\mathrm{F}$ connected in series, so the capacitance between points $\mathrm{E}$ and $\mathrm{F}$, is $\frac{1}{\mathrm{C}_{\mathrm{EF}}}=\frac{1}{6}+\frac{1}{12}=\frac{2+1}{12}=\frac{3}{12}$ $\mathrm{C}_{\mathrm{EF}}=\frac{12}{3}=4 \mu \mathrm{F}$ Capacitors between points $G$ and $D$ are connected in series, therefore, $\frac{1}{\mathrm{C}_{\mathrm{GD}}}=\frac{1}{8}+\frac{1}{4}=\frac{1+2}{8}=\frac{3}{8}$ $\mathrm{C}_{\mathrm{GD}}=\frac{8}{3} \mu \mathrm{F}$ $\therefore$ Capacitance between points $\mathrm{E}$ and $\mathrm{F}$, are connected in parallel, So Capacitors between points $\mathrm{G}$ and $\mathrm{F}$, are connected in series, therefore, $\frac{1}{\mathrm{C}_{\mathrm{GF}}}=\frac{1}{1}+\frac{1}{8}=\frac{8+1}{8}=\frac{9}{8}$ $\mathrm{C}_{\mathrm{GF}}=\frac{8}{9} \mu \mathrm{F}$ $\mathrm{C}_{\mathrm{GD}}=\frac{8}{3}+\frac{8}{9}=\frac{24+8}{9}=\frac{32}{9} \mu \mathrm{F}$ $\frac{1}{\mathrm{C}_{\mathrm{AB}}}=\frac{1}{\mathrm{C}}+\frac{1}{32 / 9}$ $\frac{1}{\mathrm{C}_{\mathrm{AB}}}=\frac{1}{\mathrm{C}}+\frac{9}{32}$ Here, $\mathrm{C}_{\mathrm{AB}}=1 \mu \mathrm{F}$ $\frac{1}{\mathrm{C}}=1-\frac{9}{32}=\frac{32-9}{32}=\frac{23}{32}$ $\mathrm{C}=\frac{32}{23} \mu \mathrm{F}$
