A 6 uF capacitor is charged by a 300
V supply. It is then disconnected
from the supply and is connected to
another uncharged 3uF capacitor. How
much electrostatic energy of the first
capacitor is lost in the form of heat and
electromagnetic radiation ?
Answers
Answer:
Capacitance of a charged capacitor, C
1
=4μF=4×10
−6
F
Supply voltage, V
1
=200V
Electrostatic energy stored in C1 is given by,
E
1
=
2
1
C
1
V
1
2
=
2
1
×4×10
−6
×(200)
2
=8×10
−2
J
Capacitance of an uncharged capacitor, C
2
=2μF=2×10
−6
F
When C
2
is connected to the circuit, the potential acquired by it is V
2
.
According to the conservation of charge, initial charge on capacitor C
1
is equal to the final charge on capacitors, C
1
and C
2
∴V
2
(C
1
+C
2
)=C
1
V
1
V
2
×(4+2)×10
−6
=4×10
−6
×200
V
2
=
3
400
V
Electrostatic energy for the combination of two capacitors is given by,
E
2
=
2
1
(C
1
+C
2
)V
2
2
=
2
1
(2+4)×10
−6
×(
3
400
)
2
=5.33×10
−2
J
Hence, amount of electrostatic energy lost by capacitor C
1
=E
1
−E
2
=0.08−0.0533=0.0267
=2.67×10
−2
J
Answer:
0.27J
Explanation:
heat is a form of energy
so as we know
E=1/2 CV^2
E=1/2 (6*10^-6(300^2))
E=(3*10^-6)(9*10^4)
E=0.27J