To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch. When the switch is closed, the battery powers two paths in parallel, one of which has a resistor of resistance R1 = 95 Ω in series with an inductor of inductance L = 1.5×10−2 H , while the other has a resistor of resistance R2 = 360 Ω . What is the current supplied by the battery at a time t = 0.4 ms after the switch is closed?

Since we have two sections, for the Inductor region there would be a current [tex]i_1[/tex]. In the case of resistance 2, it will cross a current [tex]i_2[/tex]

Defined this we proceed to obtain our equations,

For [tex]i_1[/tex],

[tex]\frac{di_1}{dt}+i_1R_1 = V[/tex]

[tex]I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})[/tex]

For [tex]i_2[/tex],

[tex]I_2R_2 =V[/tex]

[tex]I_2 = \frac{V}{R_2}[/tex]

The current in the entire battery is equivalent to,

[tex]i_t = I_1+I_2[/tex]

[tex]i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})[/tex]

Our values are,

[tex]V=1V[/tex]

[tex]R_1 = 95\Omega[/tex]

[tex]L= 1.5*10^{-2}H[/tex]

[tex]R_2 =360\Omega[/tex]

Replacing in the current for t= 0.4m/s

[tex]i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})[/tex]

[tex]i= 0.0133A[/tex]

[tex]i_1 = 0.01052A[/tex]

oladayoademosu oladayoademosu · Answer 2 · 2022-04-06T07:16:08+02:00

The current supplied by the battery at a time t = 0.4 ms after the switch is closed os 9.43 mA

The current in the series R₁ and L circuit

Since in the first parallel branch,

the resistor R₁ = 95 Ω and
inductor L = 1.5 × 10⁻² H

are in series, the current i₁ in this circuit is given by

[tex]i_{1} = \frac{V}{R_{1} } (1 - e^{-\frac{L}{R}t })[/tex]

The current on the second resistor, R₂

The current in the other second parallel branch is gotten from Ohm's law.

So, V = I₂R₂ where

V = voltage of battery = 1 V
i₂ = current in resstor and
R₂ = resistance = 360 Ω

Making i₂ subject of the formua, we have

i₂ = V/R₂

Current from battery

The current from the battery is thus i = i₂ + i₂

[tex]i = \frac{V}{R_{1} } (1 - e^{-\frac{L}{R}t }) + \frac{V}{R_{2} }[/tex]

So, the current from the battery when t = 0.4 ms = 0.4 × 10⁻³ s is gotten by substituting the values of the variables into the equation for i.

So, we have with V = 1 V

[tex]i = \frac{V}{R_{1} } (1 - e^{-\frac{L}{R}t }) + \frac{V}{R_{2} }[/tex]

[tex]i = \frac{1 V}{95 \Omega} (1 - e^{-\frac{95 \Omega}{1.5 X 10^{-2} H} X 0.4 X 10^{-3} s}) + \frac{ 1 V}{360 \Omega}\\i = \frac{1 V}{95 \Omega} (1 - e^{-\frac{38 X 10^{-3}\Omega s}{1.5 X 10^{-2} H}}) + \frac{1 V}{360 \Omega} \\i = \frac{1 V}{95 \Omega} (1 - e^{-2.533}) + \frac{1 V}{360 \Omega} \\i = \frac{1 V}{95 \Omega} (1 - 0.368) + \frac{1 V}{360 \Omega} \\i = \frac{1 V}{95 \Omega} (0.632) + \frac{1 V}{360 \Omega} \\i = 0.00665 A + 0.00278 A\\i = 0.00943 A\\i = 9.43 mA[/tex]

So, the current supplied by the battery at a time t = 0.4 ms after the switch is closed os 9.43 mA

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