Given Information:
Resistance = R = 1.9 k
Ω
Capacitance = C = 30 uF
Initial charge = 30 uC
Final charge = 5 uC
Required Information:
Time taken to reduce the capacitor's charge to 5.0 μC = ?
Answer:
t = 0.101 seconds
Explanation:
The voltage across the capacitor is given by
V = V₀e^(–t/τ)
Where τ is time constant, V₀ is the initial voltage, V is the voltage after some time t
τ = RC
τ = 1900*30x10⁻⁶
τ = 0.057 sec
The initial voltage across the capacitor was
V₀ = Q/C
V₀ = 30/30
V₀ = 1 V
Voltage to reduce the charge to 5 uF
V = 5/30
V = 0.167 V
V = V₀e^(–t/τ)
0.167 = 1*e^(–t/0.057)
take ln on both sides
ln(0.167) = ln(e^(–t/0.057))
-1.789 = -t/0.057
t = 1.789*0.057
t = 0.101 seconds
The answer is B) inverted
Answer:
m per second is the answaegnzxn
Total distance covered = 384 Km
Total time taken to travel from A to B = 8 hours [from 8 am to 4 pm, there are 8 hours]
We know, Average speed = Total Distance Travelled/ Total Time Taken
Therefore, average speed = 384 Km/8 h = 384000m/8×60×60s =(384000/28800)m/s
= 13.3 m/s
Answer is 13.3 m/s
