Respuesta :
Answer:
(a). The time constant of the circuit is 2.17.
(b). The potential difference across the capacitor at t=17.0 s is 0.0396 V.
Explanation:
Given that,
Initial potential difference = 100 V
Potential difference across the capacitor = 1.00 V
(a). We need to calculate the time constant of the circuit
Using formula of potential difference
[tex]V(t)=V_{0}e^{\dfrac{-t}{RC}}[/tex]
Put the value into the formula
[tex]1.00=100e^{\dfrac{-10.0}{RC}}[/tex]
[tex]0.01=e^{\dfrac{-10.0}{RC}}[/tex]
On taking ln
[tex]ln(0.01)=\dfrac{-10}{RC}[/tex]
[tex]RC=\dfrac{-10}{ln(0.01)}[/tex]
[tex]RC=2.17[/tex]
(b). We need to calculate the potential difference across the capacitor at t=17.0 s
Using formula again
[tex]V(17)=100e^{\dfrac{-17}{2.17}}[/tex]
[tex]V{17}=0.0396\ V[/tex]
Hence, (a). The time constant of the circuit is 2.17.
(b). The potential difference across the capacitor at t=17.0 s is 0.0396 V.
Based on the calculations, the time constant of this circuit is equal to 2.172.
Given the following data:
Initial potential difference = 100 V.
Initial time = 10 seconds.
Final time = 17 seconds.
Final potential difference = 1.00 V.
How to calculate the time constant of a circuit.
Mathematically, potential difference is given by this formula:
[tex]V(t)=V_oe^{\frac{-t}{RC} }\\\\1=100e^{\frac{-10}{RC} }\\\\\frac{1}{100} =e^{\frac{-10}{RC} }\\\\0.01=e^{\frac{-10}{RC} }\\\\ln(0.01)={\frac{-10}{RC} }\\\\-4.6051={\frac{-10}{RC} }\\\\RC=\frac{-10}{-4.6051} \\\\[/tex]
RC = 2.172.
How to calculate the potential difference.
Time, t = 17.0 seconds.
[tex]V(t)=V_oe^{\frac{-t}{RC} }\\\\V(17)=100e^{\frac{-17}{2.172} }\\\\V(17)=100e^{-7.8269} }\\\\V(17)=100 \times 0.0003989[/tex]
Potential difference = 0.03989 Volt.
Read more on potential difference here: brainly.com/question/4313738
