Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
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]).
To know more about R-C circuit, visit here:
https://brainly.com/question/14687618
#SPJ4
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
