When a capacitor , charged to a voltage , is discharged through a resistor , the current will decay exponentially according to the following equation:
i=E/R e^(-t/RC)
If E=250 V,C=160 μF and R=2100 Ω, answer the following questions:
a) What is the maximum current in the circuit?
b) After how long is the current one fourth of the maximum current?
To answer these questions, we can substitute the given values into the equation for current in a charging capacitor:
i = E/R * e^(-t/RC)
a) To find the maximum current in the circuit, we need to determine the value of the current at t = 0, which is the initial value. Let's substitute t = 0 and the given values into the equation:
i = (250V) / (2100Ω) * e^(-0 / (160μF * 2100Ω))
Simplifying, we have:
i = (250V) / (2100Ω) * e^0
Since any number raised to the power of 0 is equal to 1, we have:
i = (250V) / (2100Ω)
Calculating this, we get:
i = 0.119 A
Therefore, the maximum current in the circuit is approximately 0.119 A.
b) To find the time at which the current is one fourth of the maximum current, we need to find a value of t that satisfies the equation:
i = (1/4) * (0.119 A)
Substituting the given values into the equation:
(1/4) * (0.119 A) = (250V) / (2100Ω) * e^(-t / (160μF * 2100Ω))
Simplifying, we have:
e^(-t / (160μF * 2100Ω)) = (1/4) * (0.119 A) / (250V) * (2100Ω)
Now, we can solve for t by taking the natural logarithm (ln) of both sides of the equation:
ln(e^(-t / (160μF * 2100Ω))) = ln((1/4) * (0.119 A) / (250V) * (2100Ω))
This simplifies to:
-t / (160μF * 2100Ω) = ln((1/4) * (0.119 A) / (250V) * (2100Ω))
Now, we can isolate t by multiplying both sides of the equation by (160μF * 2100Ω):
t = -ln((1/4) * (0.119 A) / (250V) * (2100Ω)) * (160μF * 2100Ω)
Calculating this, we get:
t ≈ 0.0035 s
Therefore, the current is one fourth of the maximum current after approximately 0.0035 seconds.