Chapter 30: Q80P (page 901)

In Fig. 30-63, $R = 4.0 kΩ$ , $L = 8.0 μH$ and the ideal battery has $ε = 20 v$ . How long after switch S is closed is the current 2.0 mA?

Short Answer

Expert verified

For $1.0 \times 10^{- 9} s$ after switch S is closed, the value for the current is 2.0 mA

Step by step solution

Given

$R = 4.0 kΩ = 4.0 \times 10^{3} Ω L = 8.0 μH = 8.0 \times 10^{- 6} H ε = 20 v I = 2.0 mA = 2.0 \times 10^{- 3} A$

Understanding the concept

We have the equation for current in the loop. We rearrange the equation to find the required value.

Formula:

$I = \frac{ε}{R} (1 - e^{- \frac{t}{τ_{L}}})$

$τ_{L} = \frac{L}{R}$

Calculate how long after switch S is closed is the current 2.0 mA

We have,

$I = \frac{ε}{R} (1 - e^{- \frac{t}{τ_{L}}})$

But,

$τ_{L} = \frac{L}{R} τ_{L} = \frac{8.0 \times 10^{- 6}}{4.0 \times 10^{3}} τ_{L} = 2.0 \times 10^{- 9} s 2.0 \times 10^{3} = \frac{20}{4.0 \times 10^{3}} \times (1 - e^{\frac{- t}{2.0 \times 10^{- 9}}}) (1 - e^{\frac{- t}{2.0 \times 10^{- 9}}}) = \frac{2.0 \times 10^{- 3} \times 4.0 \times 10^{3}}{20} (1 - e^{\frac{- t}{2.0 \times 10^{- 9}}}) = 0.4 e^{\frac{- t}{2.0 \times 10^{- 9}}} = 0.4 - 1 e^{\frac{- t}{2.0 \times 10^{- 9}}} = - 0.6 e^{\frac{- t}{2.0 \times 10^{- 9}}} = 0.6$