When discharging a capacitor it will
a) loose half of its charge in equal time intervals
b) loose a third of its charge in equal time intervals
c) loose a quarter of its charge in equal time intervals
d) gain half of its charge in equal time intervals
I know that d) isn't correct. I believe it's a).
a, b and c are all correct. The time intervals are different in each case.
c = q/v
v = q/c
dv/dt = (1/c) dq/dt = -(1/c) i (negative because the discharge current is decreasing the q on the capacitor)
Now let's discharge this capacitor through a resistance R
v = i R so i = v/R
put those together
dv/dt = -(1/c) v/R
dv/dt = -(1/Rc)V
dv/v = -(1/Rc_) dt
ln v = k -1/(RC) t where k is a constant of integration
v = e^[k -(1/Rc)t ] = e^k e^-(t/Rc)
v = initial v * e^-(t/Rc)
call initial v = Vo
v = Vo e^-(t/Rc)
i = (Vo/R) e^(-t/Rc)
q = Vo c e^(-t/Rc)
So the chare is exponentially decreasing with time
If it loses half its charge in T seconds, it will lose half again in T seconds, and half again in another T
so charge versus time table would be
t = 0, q = Vo c
t = T, q = (1/2) Vo c
t = 2T, q = (1/2)(1/2) = 1/4 Vo c
t = 3 T, q = (1/2)(1/4) = (1/8) Vo c
etc, so it loses 1/2 of its MOST RECENT charge in each "half life"
NOW
exactly the same logic applies to the 1/3 rule
t = 0, q = Vo c
t = T1, q = (2/3)Vo c
t = 2 T1, q = (2/3)(2/3) = (4/9) Vo c
t = 2 T1, q = (2/3)(4/9) = (8/27) Vo c
etc
in other words since the decay of charge is exponential, it loses the same fraction of the most recent value (not the original value) in every time interval no matter if 1/2 or 1/3 or 15/33 or whatever.
To determine the correct answer, we need to understand the concept of discharging a capacitor. When a capacitor is connected to a circuit, it accumulates charge and stores energy. However, when the capacitor is disconnected from the power source and allowed to discharge, the stored charge gradually decreases over time.
To solve this question, we can analyze the discharge process by examining the relationship between the charge remaining on the capacitor and the time elapsed.
The discharge of a capacitor follows an exponential decay pattern, meaning its charge decreases exponentially over time. The equation describing this behavior is:
Q(t) = Q₀ * e^(-t/RC)
In this equation:
- Q₀ is the initial charge on the capacitor
- t represents time elapsed
- R is the resistance in the circuit
- C is the capacitance of the capacitor
We can observe that the discharge process does not result in equal time intervals for equal reductions in charge. Instead, the discharge occurs exponentially, leading to a decreasing rate of charge loss.
Looking at the answer choices:
a) Losing half of its charge in equal time intervals suggests a linear decrease, which is not correct.
b) Losing a third of its charge in equal time intervals also suggests a linear decrease, which is not correct.
c) Losing a quarter of its charge in equal time intervals suggests an exponential decrease, which is consistent with the behavior of a discharging capacitor.
Based on this analysis, we can conclude that the correct answer is c) The capacitor will lose a quarter of its charge in equal time intervals as it discharges.