1
JEE Main 2017 (Offline)
+4
-1
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:
A
2
B
16
C
32
D
24
2
JEE Main 2017 (Offline)
+4
-1
In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be:

A
$$CE{{{r_1}} \over {({r_1} + r)}}$$
B
CE
C
$$CE{{{r_1}} \over {({r_2} + r)}}$$
D
$$CE{{{r_2}} \over {(r + {r_2})}}$$
3
JEE Main 2016 (Online) 10th April Morning Slot
+4
-1
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 :

A
$${{31} \over {23}}\,\mu F$$
B
$${{32} \over {23}}\,\mu F$$
C
$${{33} \over {23}}\,\mu F$$
D
$${{34} \over {23}}\,\mu F$$
4
JEE Main 2016 (Online) 9th April Morning Slot
+4
-1
Three capacitors each of 4 $$\mu$$F are to be connected in such a way that the effective capacitance is 6 $$\mu$$F. This can be done by connecting them :
A
all in series
B
two in series and one in parallel
C
all in parallel
D
two in parallel and one in series
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