1
JEE Main 2020 (Online) 5th September Morning Slot
+4
-1
Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :
A
Zero
B
$${3 \over 2}C{V^2}$$
C
$${9 \over 2}C{V^2}$$
D
$${{25} \over 6}C{V^2}$$
2
JEE Main 2020 (Online) 4th September Evening Slot
+4
-1
A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $${C \over 2}$$. The energy loss in the process after the charge is distributed between the two capacitors is :
A
$${1 \over 2}CV_0^2$$
B
$${1 \over 4}CV_0^2$$
C
$${1 \over 3}CV_0^2$$
D
$${1 \over 6}CV_0^2$$
3
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
4
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
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