A 7.35 H inductor with negligible resistance is placed in series with a 14.1 V battery, a 3.00 ? resistor, and a switch. The switch is closed at time t = 0 seconds.
(a) Calculate the initial current at t = 0 seconds.
(b) Calculate the current as time approaches infinity.
(c) Calculate the current at a time of 6.25 s.
(d) Determine how long it takes for the current to reach half of its maximum.
Answer:
(a). The initial current at t=0 is zero.
(b). The current as time approaches infinity is 4.7 A
(c). The current at a time of 6.25 s is 4.33 A.
(d). The time is 1.69 sec.
Explanation:
Given that,
Inductance = 7.35 H
Voltage = 14.1 V
Resistance= 3.00 ohm
(a). We need to calculate the initial current at t = 0
Using formula of current
[tex]I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})[/tex]
Put the value into the formula
[tex]I=(\dfrac{E}{R})(1-e^{\dfrac{-0\timesR}{L}})[/tex]
[tex]I=(\dfrac{E}{R})(1-1)[/tex]
[tex]I=0[/tex]
(b). We need to calculate the current as time approaches infinity.
Using formula of current
[tex]I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})[/tex]
Put the value into the formula
[tex]I=(\dfrac{14.1}{3.00})(1-e^{\dfrac{-\infty\times3.00}{7.35}})[/tex]
[tex]I=\dfrac{14.1}{3.00}(1-0)[/tex]
[tex]I=4.7\ A[/tex]
(c). We need to calculate the current at a time of 6.25 s
Using formula of current
[tex]I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})[/tex]
Put the value into the formula
[tex]I=(\dfrac{14.1}{3.00})(1-e^{\dfrac{-6.25\times3.00}{7.35}})[/tex]
[tex]I=4.33\ A[/tex]
(d). We need to calculate the time
Using formula of current
[tex]I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})[/tex]
Put the value into the formula
[tex]2.35=(\dfrac{14.1}{3.00})(1-e^{\dfrac{-t\times3.00}{7.35}})[/tex]
[tex]ln(0.5)=(\dfrac{-t\times3.00}{7.35}})[/tex]
[tex]t=\dfrac{(\ln(2)\times7.35)}{3.00}[/tex]
[tex]t=1.69\ s[/tex]
Hence, (a). The initial current at t=0 is zero.
(b). The current as time approaches infinity is 4.7 A
(c). The current at a time of 6.25 s is 4.33 A.
(d). The time is 1.69 sec.