Respuesta :
a)The time constant of the circuit is 2sec.
b)The maximum charge on the capacitor is 0.18C
c)The charge on the capacitor at one time constant is 0.18×(1-(1/[tex]e^\frac{t}{500}[/tex]).
a)We have resistor resistance 100ohm,capacitor capacitance as 20mf and EMF value as 9V.
We know very well that the time constant of the circuit is defined as the product of Resistance and Capacitance, or in other words
τ=R×C
Therefore,τ=100ohm×20×10⁻³F
=>τ=2sec.
b)The maximum charge on the capacitor is defined as the product of capacitance and voltage applied by the battery.
In other words, Q=C×E
=>Q=20×10⁻³F×9V
=>Q= 180×10⁻³C
=>Q=0.18C
c)Now,we need to find the charge on the capacitor at one constant time.We know very well that charge on the capacitor at any given time is represented by
q(t)=q₀(1−1 /[tex]e\frac{t}{RC}[/tex])
where q₀ is the maximum charge induced on the capacitor
Therefore, q(t)=0.18×(1-(1/[tex]e^\frac{t}{100.5}[/tex])
=>q(t)=0.18×(1-(1/[tex]e^\frac{t}{500}[/tex])
Hence, a)time constant is 2sec,b)maximum charge is 0.18C and c)charge at one time constant on capacitor is q(t)=0.18×(1-(1/[tex]e^\frac{t}{500}[/tex]).
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