Chapter 21: Q65PE (page 777)

The duration of a photographic flash is related to an RC time constant, which is 0.100 µs for a certain camera. (a) If the resistance of the flash lamp is 0.0400 Ω during discharge, what is the size of the capacitor supplying its energy? (b) What is the time constant for charging the capacitor, if the charging resistance is 800 kΩ?

Short Answer

Expert verified

a) The size of the capacitor supplying its energy is $2.6 μ F$ .

b) The time constant for charging the capacitor, if the charging resistance is 800 kΩ is 2 s.

Step by step solution

Definition of resistance

In an electrical circuit, resistance is a measure of the resistance to current flow. The Greek letter omega (Ω) is used to represent resistance in ohms.

Calculate the capacitance

(a)

In this problem, we'll look at a flash camera with an RC time constant of.

$τ = 0.1 μ s (\frac{10^{- 6} s}{1 μ s}) = 1.0 \times 10^{- 7} s .$

If the flash's resistance is $R = 0.04 Ω$ , we'll compute the capacitance using the formula,

$τ = R C C = \frac{τ}{R} = \frac{0.1 μ s}{0.04 Ω} = 2.6 \times 10^{- 6} F (\frac{1 μ F}{10^{- 6} F}) = 2.6 μ F Therefore, the size of the capacitor supplying its energy is 2.6 μF .$

Calculate the time constant

(b)

If the resistance of the flash is $R = 800 k Ω (\frac{1000 Ω}{1 k Ω}) = 8.00 \times 10^{5} Ω$ , and the capacitance is the same, we calculate the time constant as

$τ = R C = (8.00 \times 10^{5} Ω) \times (2.6 \times 10^{- 6} F) = 2 s Hence, The time constant for charging the capacitor, if the charging resistance is 800 kΩ is 2 s$