A capacitor of capacitance 
F is first charged by connecting across 10 V battery, then it is allowed to get discharged through 2 
and 4
resistor by closing the key 
as shown in figure. The total energy dissipated in 2
resistor is equal to
.png)
a)
0.15 m J
b)
0.5 m J
c)
0.05 m J
d)
1.0 m J
F is first charged by connecting across 10 V battery, then it is allowed to get discharged through 2 and 4 resistor by closing the key as shown in figure. The total energy dissipated in 2 resistor is equal toa)
0.15 m J
b)
0.5 m J
c)
0.05 m J
d)
1.0 m J
Question ID - 150495 | SaraNextGen Top Answer A capacitor of capacitance F is first charged by connecting across 10 V battery, then it is allowed to get discharged through 2 and 4 resistor by closing the key as shown in figure. The total energy dissipated in 2 resistor is equal to
a)
0.15 m J
b)
0.5 m J
c)
0.05 m J
d)
1.0 m J
A capacitor of capacitance F is first charged by connecting across 10 V battery, then it is allowed to get discharged through 2 and 4 resistor by closing the key as shown in figure. The total energy dissipated in 2 resistor is equal to
|
a) |
0.15 m J |
b) |
0.5 m J |
c) |
0.05 m J |
d) |
1.0 m J |
1 Answer
127 votes
Answer Key / Explanation : (c) -
|
(c) Total energy stored in capacitor,
Energy dissipated in
|
J
J = 0.05 mJ127 votes
Answer Link
127