1
JEE Main 2013 (Offline)
+4
-1
Two capacitors $${C_1}$$ and $${C_2}$$ are charged to $$120$$ $$V$$ and $$200$$ $$V$$ respectively. It is found that connecting them together the potential on each one can be made zero. Then
A
$$5{C_1} = 3{C_2}$$
B
$$3{C_1} = 5{C_2}$$
C
$$3{C_1} + 5{C_2} = 0$$
D
$$9{C_1} = 4{C_2}$$
2
AIEEE 2012
+4
-1
The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant $$\tau$$ of this circuit lies between
A
100 sec and 150 sec
B
0 and 50 sec
C
50 sec and 100 sec
D
150 sec and 200 sec
3
AIEEE 2010
+4
-1
Let $$C$$ be the capacitance of a capacitor discharging through a resistor $$R.$$ Suppose $${t_1}$$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $${t_2}$$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $${t_1}/{t_2}$$ will be
A
$$1$$
B
$${1 \over 2}$$
C
$${1 \over 4}$$
D
$$2$$
4
AIEEE 2008
+4
-1
A parallel plate capacitor with air between the plates has capacitance of $$9$$ $$pF.$$ The separation between its plates is $$'d'.$$ The space between the plates has dielectric constant $${k_1}$$ $$=3$$ and thickness $${d \over 3}$$ while the other one has dielectric constant $${k_2} = 6$$ and thickness $${{2d} \over 3}$$. Capacitance of the capacitor is now
A
$$1.8$$ $$pF$$
B
$$45$$ $$pF$$
C
$$40.5$$ $$pF$$
D
$$20.25$$ $$pF$$
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