P = dE/dt
dE = P*dt
dE = 1 x 10^5 * 12 x 10^-6 = 12 J
dE = CV^2/2
C = 2dE/V^2
V = 320 V
C = 2.3 x 10^-4 F
The flash unit in a camera uses a special circuit to "step up" the 3.0 V...
The flash unit in a camera uses a special circuit to "step up" the 3.0 V from the batteries to 270 V , which charges a capacitor. The capacitor is then discharged through a flashlamp. The discharge takes 11 μs , and the average power dissipated in the flashlamp is 1.0×10^5 W . What is the capacitance of the capacitor?
The flash unit in a camera uses a special circuit to "step up" the 3.0 V from the batteries to 360 V , which charges a capacitor. The capacitor is then discharged through a flashlamp. The discharge takes 12 μs , and the average power dissipated in the flashlamp is 1.0×105 W . What is the capacitance of the capacitor? use 2 sigfigs I have tried 0.27 F and 0.26 F but so far both are wrong on mastering physics.
5. AM: Capacitors are used in many applications where you need to supply a short burst of energy, for example, a mobile defibrillator or a camera's flashlamp. The flash unit in a camera uses a special circuit to “step up” the 3.0 V from the batteries to 300 V, and the 300 V charges a capacitor (otherwise the 3.0 V battery would take very long time to charge the capacitor). The capacitor is then discharged through a flashlamp. The discharge...
The RC charging circuit in a camera flash unit has a voltage
source of 275 V and a capacitance of 125F.
Find its resistance R if the capacitor charges to 90% of its
final value in 15 s.
Find the average current delivered to the flash bulb if the
capacitor discharges 90% of its full charge in 1.00 ms.
The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of a 705 mu F capacitor is 320 V. Determine the energy that is used to produce the flash in this unit. J Assuming that the flash lasts for 5.0 times 10^-3 s, find the effective power or "wattage" of the flash. W
The RC charging circuit in a camera flash unit has a voltage source of 295 V and a capacitance of 116 µF. (a) Find its resistance R (in ohms) if the capacitor charges to 90.0% of its final value in 12.2 s. Ω (b) Find the average current (in A) delivered to the flash bulb if the capacitor discharges 90.0% of its full charge in 1.04 ms. A
1. Calculate the work done by a 8.5-V battery as it charges a 7.5-μF capacitor in the flash unit of a camera. Express your answer using two significant figures. W=?? μJ 2. Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the plates is 360 V , and the plate separation is 0.261 mm . uE= ?? Jm^3
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Problem 6 Consider the circuit shown in (Figure 1). Suppose that V, = 480_0°V (rms). Part A Find the average power dissipated in the line in the figure Express your answer three significant figures and include the appropriate units. HHA ? P = Value Units Submit Request Answer Part B Figure 1 of 1 > Find the capacitive reactance that, when connected in parallel, with the load will make the load look purely resistive. Express your answer three significant...
An ac series circuit consists of a voltage source of frequency 60 Hz and voltage amplitude V, a 358-12 resistor, and a capacitor of capacitance 6.2 uF. What must be the source voltage amplitude V for the average electrical power consumed in the resistor to be 432 W? There is no inductance in the circuit. Express your answer using two significant figures. temp@tes Symbols uado ředo reset keyboard shortcuts help, VE
Part A An ac series circuit consists of a voltage source of frequency 60 Hz and voltage amplitude V. a 354- resistor, and a capacitor of capacitance 3.3 uF. What must be the source voltage amplitude V for the average electrical power consumed in the resistor to be 468 W? There is no inductance in the circuit. Express your answer using two significant figures. VE V Submit Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining Provide Feedback...