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Problem 5: A resistor and inductor are connected to a 9.0 V battery by a switch as shown. The moment the switch is closed, current flows through the circuit. The resistor has a resistance of R = 260 ? and the inductor has an inductance of L = 110 mH. Randomized Variables R = 260 ? L = 110 mH Part (a) At time t = 0 the switch is closed and current flows through the circuit. The current increases with time and eventually reaches a steady state value of i_max. Calculate the maximum current i_max in units of milliamps. Part (b) Calculate the time constant, ?, of the circuit, in seconds. Part (c) Determine the time, in seconds, at which the current has a value of i(t_50) = 50% of i_max. Part (d) Determine the time, in seconds, at which the current has a value of i(t_99) = 99% of i_max.

          Problem 5: A resistor and inductor are connected to a 9.0 V battery by a switch as shown. The moment the switch is closed, current flows through the circuit. The resistor has a resistance of R = 260 ? and the inductor has an inductance of L = 110 mH.

Randomized Variables
R = 260 ?
L = 110 mH

Part (a) At time t = 0 the switch is closed and current flows through the circuit. The current increases with time and eventually reaches a steady state value of i_max. Calculate the maximum current i_max in units of milliamps.

Part (b) Calculate the time constant, ?, of the circuit, in seconds.

Part (c) Determine the time, in seconds, at which the current has a value of i(t_50) = 50% of i_max.

Part (d) Determine the time, in seconds, at which the current has a value of i(t_99) = 99% of i_max.

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Problem 5: A resistor and inductor are connected to a 9.0 V battery by a switch as shown. The moment the switch is closed, current flows through the circuit. The resistor has a resistance of R = 260 ? and the inductor has an inductance of L = 110 mH.

Randomized Variables
R = 260 ?
L = 110 mH

Part (a) At time t = 0 the switch is closed and current flows through the circuit. The current increases with time and eventually reaches a steady state value of imax. Calculate the maximum current imax in units of milliamps.

Part (b) Calculate the time constant, ?, of the circuit, in seconds.

Part (c) Determine the time, in seconds, at which the current has a value of i(t50) = 50% of imax.

Part (d) Determine the time, in seconds, at which the current has a value of i(t99) = 99% of imax.

Added by Taylor O.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Qbs Educator

03/29/2022

Step 1

Given: Resistance, \(R = 260 \, \Omega\) Inductance, \(L = 110 \, mH = 110 \times 10^{-3} \, H\) Battery voltage, \(V = 9.0V\) The maximum current can be calculated using the formula: \[ i_{max} = \frac{V}{R} \] Substitute the given values: \[ i_{max} = Show more…

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Problem 5: A resistor and inductor are connected to a 9.0 V battery by a switch as shown. The moment the switch is closed, current flows through the circuit. The resistor has a resistance of R = 260 ̩ and the inductor has an inductance of L = 110 mH. Randomized Variables R = 260 ̩ L = 110 mH Part (a) At time t = 0 the switch is closed and current flows through the circuit. The current increases with time and eventually reaches a steady state value of i_max. Calculate the maximum current i_max in units of milliamps. Part (b) Calculate the time constant, τ, of the circuit, in seconds. Part (c) Determine the time, in seconds, at which the current has a value of i(t_50) = 50% of i_max. Part (d) Determine the time, in seconds, at which the current has a value of i(t_99) = 99% of i_max.

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Transcript

00:01 So here we have an lr circuit, so the time constant we know, here r is equal to 260 home and l is equal to 110 multiplied by 10 to the per minus 3 handwhip.

00:15 So time constant, so the maximum current, i0, maximum current is equal to e divided by where e is the applied voltage so putting the values we get i0 is equal to 9 divided by 260 that is equal to 0 .0 0 .0 .0 .0 .0 3 .6.

01:07 Okay so this is the maximum current.

01:12 Now the time constant time constant is what time constant is equal to l.

01:22 By r l divided by r so putting the values we get 110 multiplied by 10 to the power minus 3 divided by 260 so that is equal to 4 .231 multiplied by 10 to the power minus 4 seconds okay so this is the time constant now for the third part we know the equation is i equal to i0 multiplied by 1 minus e to the power minus r by l into t for at any instant the current t i t is given by like this but t is the time so at any instant t that i t is given by this so for the first part i by i0 is 0 .5 so 0 .5 is equal to 1 minus e to the power minus 260 divided by 110 multiplied by 10 to the power minus 3 multiplied by t.

02:32 So from here we are getting loan of 0 .5 is equal to yeah from here we are going to loan of so point five is coming in the right -hand side so is equal to to the power so loan of 0 .5 is equal to 260 divided by 110 multiplied by 10 to the power minus 3 multiplied by t so from we are getting t is equal to 2 .93 multiplied by 10 to the power minus 4 seconds okay this is for the 50 % of maximum value now for similarly for the second case we have i by i0 is equal to 0 .99 that is equal to again 1 minus e to the power minus 260 divided by 110 multiplied by 10 to the power minus 3 multiplied by t...

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