Chapter 4: Q50P (page 1016)

An inductor with inductance L = 0.300 H and negligible resistance is connected to a battery, a switch S, and two resistors, $R 1 = 12.0 Ω$ and $R 2 = 16.0 Ω$ (Fig. P30.50). The battery has emf 96.0 V and negligible internal resistance. S is closed at t = 0. (a) What are the currents just after S is closed? (b) What are $i_{1}, i_{2}$ and $i_{3}$ after S has been closed a long time? (c) What is the value of t for which $i_{3}$ has half of the final value that you calculated in part (b)? (d) When has half of its final value, what are $i_{1} and i_{2}$ ?

Short Answer

Expert verified

A) The current passing through $i_{2} = i_{1} = 8 A$ and $i_{3} = 0$

b) The current flowing in $i_{1} = 14 A$

C) The value of t is 13ms

D) i3 that has half of its final value is 11A

Step by step solution

Concept of the current passing through the resistors just after S is closed

The current passing through the resistors is given as $i_{1} = i_{2} = \frac{ε}{R}$ when $i_{3} = 0$

$i_{2} = i_{1} = \frac{96 V}{12 Ω} = 8 A$

Therefore, the current passing through $i_{2} = i_{1} =$ 8A and $i_{3} = 0$

Calculate i1, i2, and i3 after S has been closed a long time

When reached steady state $|\frac{{di}_{3}}{dt}| = 0$ ,Potential across the inductor is $ε_{L} = - L |\frac{{di}_{3}}{dt}| = 0$

$i_{2} = \frac{ε}{R_{2}} = \frac{96 V}{12 Ω} = 8 A i_{3} = \frac{ε}{R_{2}} = \frac{96 V}{16 Ω} = 6 A$

By Kirchhoff’s rule,

$i_{2} = i_{3} = i_{1} 8 A + 6 A = 14 A$

Therefore, the current flowing in $i_{1} = 14 A$

Calculate the value of t for which i3 has half of the final value

$E = \frac{L}{R_{2}} In (\frac{- ε / R_{2}}{i_{3} - ε / R_{2}}) i_{3} = \frac{6 A}{2} = 3 A t = \frac{L}{R_{2}} In (\frac{- ε / R_{2}}{i_{3} - ε / R_{2}}) = \frac{0.3}{16} In (\frac{- 96 / 16}{3 - 96 / 16}) = 0.0130 s t = 13 ms$

Therefore, the value of t is $13 ms$

Calculate i3 that has half of its final value

$i_{3}$ has half of its value so, $i_{3} = \frac{6 A}{2} = 3 A$ and $i_{2} = \frac{ε}{R_{2}} = \frac{96 V}{12 Ω} = 8 A$

By Kirchhoff’s rule,

$i_{2} = i_{3} = i_{1} 8 A + 3 A = 11 A$

Therefore, i3 that has half of its final value is 11A

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Short Answer

Step by step solution

Concept of the current passing through the resistors just after S is closed

Calculate i1, i2, and i3 after S has been closed a long time

Calculate the value of t for which i3 has half of the final value

Calculate i3 that has half of its final value

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