WW Circuit with inductor 2(2 points) switch has been closed? Explain your answer.
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When the switch is closed, it means that the circuit is complete and current can flow through it. Show more…
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In the circuit shown in Fig. 30.24 the bartery and the inductor have no appreciable internal resistance and there is no current in the circuit. After the switch is closed, find the readings of the ammeter (A) and voltmeters $\left(V_{1} \text { and } V_{2}\right)$ (a) the instant after the switch is closed and (b) after the switch has been closed for a very long time. (c) Which answers in parts $(a)$ and $(b)$ would change if the inductance were 24.0 $\mathrm{mH}$ instead?
After the current in the circuit of Fig. 30.28 has reached its final, steady value with switch $S_{1}$ closed and $S_{2}$ open, switch $S_{2}$ is closed, thus short circuiting the inductor. (Switch $S_{1}$ remains closed. See Problem 30.69 for numerical values of the circuit elements. (a) Just after $S_{2}$ is closed, what are $v_{a c}$ and $v_{c b}$ and what are the currents through $R_{0}, R,$ and $S_{2} ?\left(\text { b ) A long time after } S_{2} \text { is }\right.$ closed, what are $v_{a c}$ and $v_{c b}$ and what are the currents through $R_{0}$ . $R,$ and $S_{2} ?(c)$ Derive expressions for the currents through $R_{0}, R,$ and $S_{2}$ as functions of the time $t$ that has elapsed since $S_{2}$ was closed. Your results should agree with part (a) when $t=0$ and with part (b) when $t \rightarrow \infty$ . Graph these three currents versus time.
After the current in the circuit has reached its final steady value with switch S1 closed and S2 open, the switch S2 is closed, thus short-circuiting the inductor. (Switch S1 remains closed.) Let the EMF be ε = 40.0 V, the resistors are R0 = 50.0 Ω and R = 150 Ω, and the inductance is L = 4.50 H. Just after S2 is closed, what is the current through R? Express your answer in amperes.
Supratim P.
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