Question
In Fig. $27.23 a$, take $R=2.5 \mathrm{k} \Omega$ and $\mathcal{E}_{0}=50 \mathrm{~V}$. When the switch is closed, the current through the inductor rises to $10 \mathrm{~mA}$ in $30 \mu$. Find (a) the inductance and (b) the current in the circuit after many time constants.
Step 1
We can use the formula for the growth of current in an LR circuit, which is given by $I = I_0(1 - e^{-t/\tau})$, where $I_0$ is the maximum current in the circuit, $t$ is the time, and $\tau$ is the time constant of the circuit. Show more…
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In Fig. $27.23 a,$ take $R=2.5 \mathrm{k} \Omega$ and $\mathcal{E}_{0}=50 \mathrm{V} .$ When the switch is closed, the current through the inductor rises to $10 \mathrm{mA}$ in $30 \mu$ s. Find (a) the inductance and (b) the current in the circuit after many time constants.
In Fig. $27.26,$ take $\mathcal{E}_{0}=12 \mathrm{V}, R=2.7 \Omega,$ and $L=20 \mathrm{H} .$ Initially the switch is in position $B$ and there's no current anywhere. At $t=0$ the switch is thrown to position $A,$ and at $t=10 \mathrm{s}$ it's returned to $B$. Find the inductor current at (a) $t=5.0 \mathrm{s}$ and (b) $t=15 \mathrm{s}.$
A $25-\mathrm{mH}$ inductor, an $8.0-\Omega$ resistor, and a $6.0-\mathrm{V}$ battery are connected in series as in Figure $\mathrm{P} 20.43 .$ The switch is closed at $t=0 .$ Find the voltage drop across the resistor (a) at $t=0$ and (b) after one time constant has passed. Also, find the voltage drop across the inductor $(\mathrm{c})$ at $l=0$ and $(\mathrm{d})$ after one time constant has elapsed.
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