A 11.00-V battery is connected through a switch to two identical resistors and a

Question

A 11.00-V battery is connected through a switch to two identical resistors and an ideal inductor, as shown in the figure. Each of the resistors has a resistance of 181.0 , and the inductor has an inductance of 4.00 H. The switch is initially open.

a) Immediately after the switch is closed, what is the current in resistor R1 and in resistor R2?

b) At 50 ms after the switch is closed, what is the current in resistor R1 and in resistor R2?

c) At 500 ms after the switch is closed, what is the current in resistor R1 and in resistor R2?

Explanation / Answer

To answer this question we need to know the internal resistance of the battery (Rb), and we are not told that; perhaps we are intended to assume that it is zero. If so, remaind your teacher that there is a circuit symbol for an ideal voltage generator which can be used to give a precise indication of such a meaning. The symbol is simply a circle with the direction and magnitude of the emf drawn beside it. b) current in R1 = 11/(181 + Rb) = 0.06077A (if Rb is taken to be 0) The current (I) in the inductor is given by I = (V/R)*(1 - e^(-t*R/L)) where R = 181 + Rb after 50ms, I = .0544mA with Rb = 0 c) after 500ms I = .06077mA

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