Question

This circuit consists of an ε=6.00V battery, and r=50.5kΩ resistor, a .c=132μF capacitor, a flashbulb, and two switches. Initially, the capacitor is uncharged, and the two switches are open. To charge the unit, switch s1 is closed; to fire the flash, switch s2 (which is connected to the camera's shutter) is closed. How long does it take to charge the capacitor to 4.00V?

a. 0.139 s
b. 0.278 s
c. 0.416 s
d. 0.554 s

Answer 1

To determine the time required to charge a capacitor to 4.00V in a specified circuit, use the formula for an RC circuit's capacitor charging and solve for time. The calculation gives an approximate time of 0.278 seconds, matching with answer choice (b).

Step-by-step explanation:

To find out how long it takes to charge the capacitor to 4.00V in a circuit consisting of a 6.00V battery, a 50.5kΩ resistor, and a 132μF capacitor, we use the formula for the charging of a capacitor in an RC circuit:

V(t) = V_0 (1 - e^{-t/(RC)})

Where V(t) is the voltage across the capacitor at time t, V_0 is the final voltage across the capacitor (6.00V in this case), R is the resistance (50.5kΩ), C is the capacitance (132μF), and e is the natural logarithm base.

Rearrange the formula to solve for t:

t = -RC ln(1 - V(t)/V_0)

Plugging in the values:

R = 50.5 x 10^3 Ω

C = 132 x 10^{-6} F

t = -(50.5 x 10^3)(132 x 10^{-6}) ln(1 - 4.00/6.00)

The answer is calculated to be approximately 0.278 seconds, which correlates with answer choice (b).