# 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 close

Question

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.

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1 year 2021-09-03T17:09:44+00:00 1 Answers 12 views 0

(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

$$I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})$$

Put the value into the formula

$$I=(\dfrac{E}{R})(1-e^{\dfrac{-0\timesR}{L}})$$

$$I=(\dfrac{E}{R})(1-1)$$

$$I=0$$

(b). We need to calculate the current as time approaches infinity.

Using formula of current

$$I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})$$

Put the value into the formula

$$I=(\dfrac{14.1}{3.00})(1-e^{\dfrac{-\infty\times3.00}{7.35}})$$

$$I=\dfrac{14.1}{3.00}(1-0)$$

$$I=4.7\ A$$

(c).  We need to calculate the current at a time of 6.25 s

Using formula of current

$$I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})$$

Put the value into the formula

$$I=(\dfrac{14.1}{3.00})(1-e^{\dfrac{-6.25\times3.00}{7.35}})$$

$$I=4.33\ A$$

(d). We need to calculate the time

Using  formula of current

$$I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})$$

Put the value into the formula

$$2.35=(\dfrac{14.1}{3.00})(1-e^{\dfrac{-t\times3.00}{7.35}})$$

$$ln(0.5)=(\dfrac{-t\times3.00}{7.35}})$$

$$t=\dfrac{(\ln(2)\times7.35)}{3.00}$$

$$t=1.69\ s$$

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.