Capacitors are used in many applications where you need to supply a short burst of energy. A 143 μF capacitor in an electronic flash lamp supplies an average power of 12.6 kW to the lamp for 2.95 ms. To what potential difference must the capacitor initially be charged?
What is its initial charge?
To determine the potential difference the capacitor must initially be charged to, we can use the formula for average power in a capacitor:
P = (1/2) * C * (V^2) * f
where:
P is the average power in watts
C is the capacitance in farads
V is the potential difference in volts
f is the frequency in hertz
From the given information, we know:
P = 12.6 kW = 12,600 W
C = 143 μF = 143 × 10^(-6) F
f = 1 / (2.95 ms) = 339.03 Hz
Substituting these values into the formula, we can solve for V:
12,600 = (1/2) * (143 × 10^(-6)) * (V^2) * 339.03
Simplifying the equation:
12,600 = 0.000071 * (V^2) * 339.03
V^2 = (12,600 / (0.000071 * 339.03))
V^2 = 5147.687
V = √5147.687
V ≈ 71.77 V
Therefore, the potential difference the capacitor must initially be charged to is approximately 71.77 volts.
To find the initial charge of the capacitor, you can use the formula:
Q = C * V
where:
Q is the charge in coulombs
C is the capacitance in farads
V is the potential difference in volts
Therefore, the initial charge of the capacitor is:
Q = (143 × 10^(-6)) F * 71.77 V
Q ≈ 0.01028 C
So, the initial charge of the capacitor is approximately 0.01028 coulombs.
To find the potential difference that the capacitor must initially be charged to, we can use the formula for the average power supplied by a capacitor:
P = (1/2) * C * V^2,
where P is the average power, C is the capacitance, and V is the potential difference.
Given that the average power is 12.6 kW (or 12,600 W), and the capacitance is 143 μF (or 143 * 10^-6 F), we need to find V.
Rearranging the formula, we have:
V^2 = (2 * P) / C,
V^2 = (2 * 12,600) / (143 * 10^-6),
V^2 = 1.757 * 10^11,
V ≈ 4.19 * 10^5 V.
Therefore, the potential difference that the capacitor must initially be charged to is approximately 4.19 * 10^5 V.
To find the initial charge of the capacitor, we can use the formula:
Q = C * V,
where Q is the charge, C is the capacitance, and V is the potential difference.
Using the given values of C = 143 μF and V ≈ 4.19 * 10^5 V, we can calculate Q:
Q = (143 * 10^-6) * (4.19 * 10^5),
Q = 60.017 C (approx).
So, the initial charge of the capacitor is approximately 60.017 Coulombs.