1
MHT CET 2021 24th September Morning Shift
+1
-0

Two identical capacitors have the same capacitance '$$\mathrm{C}$$'. One of them is charged to potential '$$\mathrm{V_1}$$' and the other to $$\mathrm{V_2}$$. The negative ends of the capacitors are connected together. When positive ends are also connected, the decrease in energy of the combined system is

A
$$\frac{1}{4} C\left(V_1-V_2\right)^2$$
B
$$\frac{1}{2} C\left(v_1^2+v_2^2\right)$$
C
$$\frac{1}{2} C\left(V_1^2-V_2^2\right)$$
D
$$\frac{1}{2} C\left(V_1+V_2\right)^2$$
2
MHT CET 2021 23rd September Evening Shift
+1
-0

If the potential difference across a capacitor is increased from $$5 \mathrm{~V}$$ to $$15 \mathrm{~V}$$, then the ratio of final energy to initial energy stored in the capacitor is

A
$$1: 3$$
B
$$27: 1$$
C
$$3: 1$$
D
$$9: 1$$
3
MHT CET 2021 23rd September Evening Shift
+1
-0

The charge on each capacitor when a voltage source os 15 V is connected in the circuit as shown, is

A
75 $$\mu$$C
B
150 $$\mu$$C
C
30 $$\mu$$C
D
60 $$\mu$$C
4
MHT CET 2021 23th September Morning Shift
+1
-0

Two parallel plates with dielectric placed between the plates are as shown in figure. The resultant capacity of capacitor will be [A = area of plate. $$t_1, t_2$$ and $$t_3$$ are thickness of dielectric slabs, $$\mathrm{k}_1, \mathrm{k}_2$$ and $$\mathrm{k}_3$$ are dielectric constants.

A
$$\frac{A \varepsilon_0}{\left[\frac{t_1+t_2+t_3}{k_1+k_2+k_3}\right]}$$
B
$$\frac{A \varepsilon_0\left(k_1 k_2 k_3\right)}{t_1 t_2 t_3}$$
C
$$A \varepsilon_0\left[\frac{k_1}{t_1}+\frac{k_2}{t_2}+\frac{k_3}{t_3}\right]$$
D
$$\frac{A \varepsilon_0}{\left[\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}\right]}$$
EXAM MAP
Medical
NEET