Physics, 10.10.2019 03:30 mya6137

Three resistors (r1 = 120 ohms, r2 = 330 ohms, and r3 = 240 ohms) and an ideal inductor (l = 1.6 mh) are connected to a battery (v = 9 v) through a switch as shown in the figure below.

the switch has been open for a long time before it is closed at t = 0. at what time t0, does the current through the inductor (i3) reach a value that is 63% of its maximum value?

Three resistors (r1 = 120 ohms, r2 = 330 ohms, and r3 = 240 ohms) and an ideal inductor (l = 1.6 mh)

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Answer from: kyleescott9525

The time at which the current through the inductor reaches 63% of the maximum current is $\fbox{\begin\\4.85 \mu s\end{minispace}}$ or $\fbox{\begin\\4.85 \times {10^{ - 6}}\,{\text{s}}\end{minispace}}$ .

Further Explanation:

At $t = 0\,{\text{s}}$ there is no current in the circuit because the switch is not closed and the circuit is not complete. The current across the LR circuit increases exponentially, when switch is closed, and becomes steady after certain time.

Given:

The value of resistor is $120\,\Omega$ .

The value of resistor is $330\,\Omega$ .

The value of resistor is $240\,\Omega$ .

The value of the inductor is $1.6\,{\text{mH}}$ .

The voltage applied across the circuit is $9\,{\text{V}}$ .

Concept:

To determine the value of effective resistance of this circuit we need to look at the circuit from inductor’s side i.e., from inductor’s side the resistors ${R_3}$ is connected in series with the parallel combination of resistors ${R_1}$ and ${R_2}$ .

The effective resistance of the circuit is:

$\fbox{\begin\\{R_{eff}} = {R_3} + \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}\end{minispace}}$ …… (1)

Here, ${R_{eff}}$ is the effective resistance of the circuit.

Substitute the $120\,\Omega$ for ${R_1}$ , $330\,\Omega$ for ${R_2}$ and $240\,\Omega$ for ${R_3}$ in equation (1).

$\begin{aligned}{R_{eff}}&=240\,\Omega+\frac{{\left( {120\,\Omega } \right) \times 330\,\Omega }}{{120\,\Omega +330\,\Omega }} \\&=328\,\Omega\\ \end{aligned}$

The current through the inductor is:

$\fbox{\begin\\i = {i_0}\left( {1 - {e^{ - \frac{{t{R_{eff}}}}{L}}}} \right)\end{minispace}}$ ...... (2)

Here, is the current across the inductor, ${i_0}$ is the maximum current in the circuit and is the inductance across the inductor.

The current across the inductor is equal to the 63% or times of the maximum current in the circuit.

The current across the inductor is:

$i = 0.63{i_0}$

Substitute $0.63{i_0}$ for , $328 \Omega$ for ${R_{eff}}$ and $1.6\,{\text{mH}}$ for in equation (2).

$0.63{i_0} = {i_0}\left( {1 - {e^{ - \frac{{t\left( {328\Omega } \right)}}{{\left( {1.6\,{\text{mH}}} \right)}}}}} \right)$

Simplify the above expression.

${e^{ - \left( {2.05 \times {{10}^6}} \right)t}}= 0.37$

Taking natural log on both sides and simplify.

$\begin{aligned}t&=4.85\, \times {10^{ - 6\,}}\,{\text{s}} \\&=4.85\mu \text{s}}\\\end{aligned}$

Thus, the time at which the current through the inductor reaches 63% of the maximum current is $\fbox{\begin\\4.85 \mu s\end{minispace}}$ or $\fbox{\begin\\4.85 \times {10^{ - 6}}\,{\text{s}}\end{minispace}}$ .

Learn more:

1. Conservation of energy

2. Average translational energy

3. The motion of a body under friction

Answer Details:

Grade: Middle School

Subject: Physics

Chapter: Current Electricity

Keywords:

Resistor circuit, LR circuit, current, current across inductor, time constant, 4.85 microsecond, 4.85 microsec, 4.85 micros, 4.85*10-6 s, 4.85*10^6 s, 4.85*10-6 sec, 4.85*10^6 sec.