Chapter 4: Q23E (page 1014)

An inductor with an inductance of $2.5 H$ and a resistance of $8 Ω$ is connected to the terminals of a battery with an emf of $6 V$ and negligible internal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current islocalid="1664185519628" $0.25 s$ ; (c) the current $0.5 A$ after the circuit is closed; (d) the final steady-state current.

Expert verified

a) $\frac{d i}{d t} = 2.4 A / s$

b) $\frac{d i}{d t} = 0.8 A / s$

c)role="math" localid="1664185700617" $i = 0.413 A$

d) $i = 0.75 A$

Step by step solution

The formula for the current in an RL circuit is $i = \frac{ε}{R} (1 - e^{- \frac{R t}{L}})$ .

At $t = 1$

$i = \frac{ε}{R} (1 - e^{0}) i = \frac{ε}{R} (1 - 1) i = 0$

Initially the current is zero. The rate of increase of current is given by

$\frac{d i}{d t} = \frac{ε}{L} \frac{d i}{d t} = \frac{6}{2.5} \frac{d i}{d t} = 2.4 A / s$

The rate of current at any instant is given by $\frac{d i}{d t} = \frac{ε - i R}{L}$ .

$\frac{d i}{d t} = \frac{6 - (0.5 \times 8)}{2.5} \frac{d i}{d t} = 0.8 A / s$

$i = \frac{ε}{R} (1 - e^{- \frac{R t}{L}}) i = \frac{6}{8} (1 - e^{- \frac{8 \times 025}{2.5}}) i = 0.75 (1 - e^{- 0.8}) i = 0.413 A$

At the infinite time, $e^{- \infty} \to 0$ . The steady state current will be $i = \frac{ε}{R}$ .

$i = \frac{6}{8} i = 0.75 A$

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