JEE MAIN - Physics (2017 (Offline) - No. 11)
A capacitance of 2 $$\mu $$F is required in an electrical circuit across a potential difference of 1.0 kV. A large
number of 1 $$\mu $$F capacitors are available which can withstand a potential difference of not more than 300 V.
The minimum number of capacitors required to achieve this is:
2
16
32
24
Explanation
To get a capacitance of 2 μF arrangement of capacitors of capacitance 1μF as shown in figure
8 capacitors of 1μF in parallel with four such branches in series i.e., 32 such capacitors are
required.
_en_11_2.png)
$${1 \over {{C_{eq}}}} = {1 \over 8} + {1 \over 8} + {1 \over 8} + {1 \over 8}$$
$$ \Rightarrow $$ $${1 \over {{C_{eq}}}} = {1 \over 2}$$
$$ \Rightarrow $$ $${{C_{eq}} = 2}$$
$${1 \over {{C_{eq}}}} = {1 \over 8} + {1 \over 8} + {1 \over 8} + {1 \over 8}$$
$$ \Rightarrow $$ $${1 \over {{C_{eq}}}} = {1 \over 2}$$
$$ \Rightarrow $$ $${{C_{eq}} = 2}$$
Comments (0)
