What voltage is applied to the 5.00 µF capacitor of a heart defibrillator that stores 35.0 J of energy?
In kv.
energy=1/2 CV^2 solve for V
11e6
To find the voltage applied to the 5.00 µF capacitor, we need to use the formula for energy stored in a capacitor.
The formula is:
E = (1/2) * C * V^2
where E is the energy stored, C is the capacitance, and V is the voltage.
We know that the energy stored is 35.0 J and the capacitance is 5.00 µF. Let's rearrange the formula and solve for V:
35.0 J = (1/2) * (5.00 µF) * V^2
First, let's convert the capacitance from µF to F:
5.00 µF = 5.00 * 10^(-6) F
Now, let's rearrange the formula and solve for V:
35.0 J = (1/2) * (5.00 * 10^(-6) F) * V^2
Multiplying both sides of the equation by 2 and rearranging:
70.0 J = (5.00 * 10^(-6) F) * V^2
Dividing both sides of the equation by (5.00 * 10^(-6)) F:
V^2 = (70.0 J) / (5.00 * 10^(-6) F)
V^2 = 1.40 * 10^(10) V^2
Taking the square root of both sides:
V = sqrt(1.40 * 10^(10) V^2)
V = 3.74 * 10^(5) V
Therefore, the voltage applied to the 5.00 µF capacitor is approximately 3.74 * 10^(5) V, or 374 kV.
To determine the voltage applied to the capacitor of a heart defibrillator, we can use the formula:
Energy (Joules) = 1/2 * Capacitance (Farads) * Voltage^2
In this case, we are given the energy stored in the capacitor (35.0 J) and the capacitance (5.00 µF). We need to solve for the voltage.
Rearranging the formula, we have:
Voltage^2 = (2 * Energy) / Capacitance
Plugging in the given values:
Voltage^2 = (2 * 35.0 J) / (5.00 × 10^-6 F)
Voltage^2 = 14000 J/F
Now, to calculate the voltage, we take the square root of both sides:
Voltage = sqrt(14000 J/F)
Converting this voltage to kilovolts (kV), we divide by 1000:
Voltage (kV) = sqrt(14000 J/F) / 1000
Now we can calculate the voltage using a calculator or by performing the calculations step by step.