Define (i) potential difference, (ii) capacitance.
A capacitor stores energy.
Describe an experiment to demonstrate that a capacitor stores energy. The ability of ... show full transcript
Worked Solution & Example Answer:Define (i) potential difference, (ii) capacitance - Leaving Cert Physics - Question 9 - 2009
Step 1
(i) the charge stored on each plate of the capacitor;
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To calculate the charge (q) stored on each plate of the capacitor, use the formula:
q=CV
Where:
C is the capacitance in farads (F)
V is the potential difference in volts (V)
Given:
C = 64 imes 10^{-6} F (converted from μF)
V = 2500 V
Substituting the values:
q=(64imes10−6)imes2500
Calculating that:
q=0.16C
Step 2
(ii) the energy stored in the capacitor;
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The energy (E) stored in the capacitor can be calculated using the formula:
E=21CV2
Substituting the values:
E=21×(64×10−6)×(2500)2
Calculating:
E=200J
Step 3
(iii) the average current that flows through the victim when the capacitor discharges in a time of 10 ms;
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find the average current (I), use the formula:
I=tq
Where:
q = charge (0.16 C from part (i))
t = time in seconds (10 ms = 0.01 s)
Substituting the values:
I=0.010.16
Calculating:
I=16A
Step 4
(iv) the average power generated as the capacitor discharges;
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The average power (P) can be determined using:
P=tE
Where:
E = energy (200 J from part (ii))
t = time (10 ms = 0.01 s)
Substituting the values:
P=0.01200
Calculating:
P=20000W
Join the Leaving Cert students using SimpleStudy...
