The magnetic field in the region shown below is increasing by 5 T/s. The circuit shown (with R=25 , C=55 µF, dimensions 1 m by 2 m) is inserted into the region as shown. The capacitor is initially uncharged.
1. How long will it take the capacitor to charge to 25% of its final charge?
Please help I tried T=RC but not right formula!
I do not have a picture of your circuit so find this hard to follow. However for an RC circuit at initial voltage V when the switch is closed:
V = i R + (1/C) integral i dt
with i = 0 at t = 0
0 = R di/dt + i/C
0 = RC di/dt + i
let i = Io (e^kt)
di/dt = k Io e^kt
0 = RC kIo e^kt + i
so
i = -RC k Io e^kt = Io e^kt
so k = -1/RC
so if in form i = Io e^-t/T
T indeed = RC
Now
The voltage around this loop = rate of change of magnetic flux
E = 5*2square meters = 10 volts
so we put a ten volt force on our RC circuit
i = Io e^-t/RC
Initially all the voltage appears on the resistor because there is no charge yet on the capacitor
Io = E/R = 10/R
so
i = (10/R)e^-t/RC
charge on capacitor Q= integral i dt
= (10 /R) (-RC e^-t/RC + k)
when t = 0, Q = 0 so
k = RC
so
Q = 10 C (1 - e^-t/RC)
Q final = 10 C
when does Q = (10/4)C ?
.25 = 1 - e^-t/RC
e^-t/RC = .75
-t/RC = ln .75 = -.2877
so
t = .2877 R C
To determine how long it will take the capacitor to charge to 25% of its final charge, we need to use the formula for the charging time of a capacitor in an RC circuit. The formula you mentioned, T = RC, will give you the total charging time for the capacitor to reach its maximum charge. However, we need to find the time it takes for the capacitor to reach 25% of its final charge, which is different.
The formula we need to use is:
t = -ln(1 - Q/Qf) × (R × C)
Where:
t is the time it takes for the capacitor to reach a certain charge (in this case, 25% of its final charge)
Q is the current charge of the capacitor at time t
Qf is the final charge of the capacitor (fully charged)
R is the resistance in the circuit
C is the capacitance of the capacitor
Since the question does not provide the final charge, we can assume that the capacitor is fully charged when the voltage across it reaches the maximum value, which is equal to the voltage of the magnetic field in the region.
Now let's calculate the time it takes for the capacitor to reach 25% of its final charge:
1. Calculate the final charge:
The magnetic field is increasing at a rate of 5 T/s. The voltage across the capacitor is equal to the rate of change of the magnetic field multiplied by the area enclosed by the circuit.
Voltage (V) = (Rate of change of magnetic field) × (Area)
V = (5 T/s) × (1 m × 2 m)
V = 10 V
Since the capacitor is initially uncharged, the final charge (Qf) is equal to the capacitance (C) multiplied by the final voltage (V).
Qf = C × V
Qf = (55 µF) × (10 V)
Qf = 550 µC
2. Calculate the time it takes for the capacitor to reach 25% of its final charge:
We want to find the time (t) when the capacitor is at 25% of its final charge, which is Q = 0.25 × Qf.
Q = (0.25) × (550 µC)
Q = 137.5 µC
Now use the formula:
t = -ln(1 - Q/Qf) × (R × C)
t = -ln(1 - 137.5 µC / 550 µC) × (25 Ω × 55 µF)
t ≈ -ln(0.75) × (25 Ω × 55 µF)
t ≈ 0.287 × (25 Ω × 55 µF)
t ≈ 396.65 ms
Therefore, it will take approximately 396.65 milliseconds (or 0.397 seconds) for the capacitor to charge to 25% of its final charge.