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 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
$${{34} \over {23}}\,\mu F$$
3
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
4
JEE Main 2016 (Offline)
+4
-1
A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge $$Q$$ (having a charge equal to the sum of the charges on the $$4$$ $$\mu \,F$$ and $$9$$ $$\mu \,F$$ capacitors), at a point distance $$30$$ $$m$$ from it, would equal :

A
$$420N/C$$
B
$$480N/C$$
C
$$240N/C$$
D
$$360N/C$$
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