0.2F capacitor is charged to 600V by a battery. On removing the battery.It is connected with another parallel plate capacitorof 1F . the potential decrease to?
Answers
Answered by
116
Hey dear,
● Answer-
V2 = 100 V
● Explanation-
# Given-
C1 = 0.2 F
C2 = 1 F
V1 = 600 V
# Solution-
With battery in place,
Q1 = C1V1
Q1 = 0.2 × 600
Q1 = 120 C
Later, capacitors are connected in parallel -
C = C1+C2
C = 0.1+1
C = 1.2 F
With battery removed,
Q2 = CV2
Q2 = 1.2V2
By law of conservation of charge -
Q1 = Q2
120 = 1.2V2
V2 = 100 V
Therefore, potential is later reduced to 100V.
Hope this helps...
● Answer-
V2 = 100 V
● Explanation-
# Given-
C1 = 0.2 F
C2 = 1 F
V1 = 600 V
# Solution-
With battery in place,
Q1 = C1V1
Q1 = 0.2 × 600
Q1 = 120 C
Later, capacitors are connected in parallel -
C = C1+C2
C = 0.1+1
C = 1.2 F
With battery removed,
Q2 = CV2
Q2 = 1.2V2
By law of conservation of charge -
Q1 = Q2
120 = 1.2V2
V2 = 100 V
Therefore, potential is later reduced to 100V.
Hope this helps...
Answered by
37
Answer:
100V
Explanation:
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