1
JEE Main 2020 (Online) 3rd September Morning Slot
+4
-1
In the circuit shown in the figure, the total charge is 750 $$\mu$$C and the voltage across capacitor C2 is 20 V. Then the charge on capacitor C2 is :
A
160 $$\mu$$C
B
450 $$\mu$$C
C
590 $$\mu$$C
D
650 $$\mu$$C
2
JEE Main 2020 (Online) 2nd September Evening Slot
+4
-1
A 10 $$\mu$$F capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is :
A
20 $$\mu$$F
B
15 $$\mu$$F
C
10 $$\mu$$F
D
30 $$\mu$$F
3
JEE Main 2020 (Online) 8th January Evening Slot
+4
-1
A capacitor is made of two square plates each of side 'a' making a very small angle $$\alpha$$ between them, as shown in figure. The capacitance will be close to :
A
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 + {{\alpha a} \over {d}}} \right)$$
B
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {4d}}} \right)$$
C
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {2d}}} \right)$$
D
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{3\alpha a} \over {2d}}} \right)$$
4
JEE Main 2020 (Online) 8th January Morning Slot
+4
-1
Effective capacitance of parallel combination of two capacitors C1 and C2 is 10 μF. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor C2 is 4 times that of C1. If these capacitors are connected in series, their effective capacitance will be :
A
4.2 μF
B
8.4 μF
C
1.6 μF
D
3.2 μF
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