JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 24)
Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be 1 $$\mu $$F is :
_10th_April_Morning_Slot_en_24_1.png)
$${{31} \over {23}}\,\mu F$$
$${{32} \over {23}}\,\mu F$$
$${{33} \over {23}}\,\mu F$$
$${{34} \over {23}}\,\mu F$$
Explanation
Equivalent capacitance of 6 $$\mu $$F and 12 $$\mu $$F is = $${{6 \times 12} \over {12 + 6}}$$ = 4 $$\mu $$F
Equivalent capacitance of 4$$\mu $$F and 4$$\mu $$F
is = 4 + 4 = 8 $$\mu $$F
New circuit is $$ \to $$
_10th_April_Morning_Slot_en_24_2.png)
equivalent capacitance of 1 $$\mu $$F and 8 $$\mu $$F is
= $${{1 \times 8} \over {8 + 1}}$$ = $${8 \over 9}$$ $$\mu $$F
Equivalent capacitance of 8 $$\mu $$F and 4 $$\mu $$F is
= $${{8 \times 4} \over {12}}$$ = $${8 \over 3}$$ $$\mu $$F
New circuit is $$ \to $$
_10th_April_Morning_Slot_en_24_3.png)
Equation capacitance of $${8 \over 9}$$ $$\mu $$F and $${8 \over 3}$$ $$\mu $$F is
= $${8 \over 9}$$ + $${8 \over 3}$$ = $${32 \over 9}$$ $$\mu $$F
$$ \therefore $$ Equivalent capacitance of AB is
CAB = $${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$$
Given that,
CAB = 1 $$\mu $$F
$$ \therefore $$ 1 = $${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$$
$$ \Rightarrow $$ C + $${{{32} \over 9}}$$ = $${{32C} \over 9}$$
$$ \Rightarrow $$ $${{23C} \over 9} = {{32} \over 9}$$
$$ \Rightarrow $$ C = $${{32} \over {23}}$$ $$\mu $$F
Equivalent capacitance of 4$$\mu $$F and 4$$\mu $$F
is = 4 + 4 = 8 $$\mu $$F
New circuit is $$ \to $$
equivalent capacitance of 1 $$\mu $$F and 8 $$\mu $$F is
= $${{1 \times 8} \over {8 + 1}}$$ = $${8 \over 9}$$ $$\mu $$F
Equivalent capacitance of 8 $$\mu $$F and 4 $$\mu $$F is
= $${{8 \times 4} \over {12}}$$ = $${8 \over 3}$$ $$\mu $$F
New circuit is $$ \to $$
Equation capacitance of $${8 \over 9}$$ $$\mu $$F and $${8 \over 3}$$ $$\mu $$F is
= $${8 \over 9}$$ + $${8 \over 3}$$ = $${32 \over 9}$$ $$\mu $$F
$$ \therefore $$ Equivalent capacitance of AB is
CAB = $${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$$
Given that,
CAB = 1 $$\mu $$F
$$ \therefore $$ 1 = $${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$$
$$ \Rightarrow $$ C + $${{{32} \over 9}}$$ = $${{32C} \over 9}$$
$$ \Rightarrow $$ $${{23C} \over 9} = {{32} \over 9}$$
$$ \Rightarrow $$ C = $${{32} \over {23}}$$ $$\mu $$F
Comments (0)
