Common Circuit

A common circuit is composed of a resistor R, an inductor L, and a source of electromotive force g. A resistor is a device used to increase the difficulty with which electricity can move through a circuit; the units of resistance are given in ohms (e). An inductor is a device that opposes the change in flow of electricity in a circuit; the units of inductance are given in henries (H). Electromotive force (emf) is another name for potential or voltage and has units of volts (V). A common source of emf is a battery. Symbols for these three circuit elements are given in the figure below. The three elements are combined in what is called a series circuit.

The flow of electricity in a circuit is called current (I) and the measure of how much work a given amount of current can do in a fixed amount of time is called potential or voltage (V). The current, I_L which is a function of time, flowing through the inductor after the switch is closed is given by

I_L(t) = (g/R)(1 - e^-tR/L).

The voltage, V_L, as a function of time across the inductor after the switch is closed is given by

V_L(t) = g e^-tR/L .

The value epsilon (g) is the electromotive force provided by the battery. For the questions below assume the battery provides six volts, the resistor has a value of one hundred ohms, and the inductor has a value of two hundred henries.

a. Verify that the quantity tR/L is unitless. An ohm is equivalent to (kg m²⁾/(sec³ A²⁾ and a henry is equivalent to (kg m²)/(sec² A²⁾ where A stands for amperes.

b. Substitute the given values for g, R, and L into the formula for I_L(t). Then graph this equation on graph paper with t as the independent variable and I_L as the dependent variable. Start your graph with t = 0 and make sure to plot enough points to get a good overall picture of the graph.

c. Substitute the given values for g, R, and L into the formula for V_L(t). On a separate piece of graph paper, graph this equation with t as the independent variable and V_L as the dependent variable. Start your graph with t = 0 and make sure to plot enough points to get a good overall picture of the graph.

d. How long before the voltage is only one volt? only 1% of the battery voltage? Compute these answers algebraically using logarithms. Then mark these times on your graph from part b.

e. What is the maximum value of the current? Estimate this answer from your graph from part b. How long before the current reaches half this value? Compute this answer algebraically using logarithms. Then mark this time on your graph from part b.

f. Power is a measure of how fast energy is converted from one form to another. For electrical circuits the power is the product of the current and the voltage. Algebraically find the power in the inductor L as a function of time. On another piece of graph paper, graph this new equation with t as the independent variable and P as the dependent variable. Start your graph with t = 0 and make sure to plot enough points to get a good overall picture of the graph.

g. Energy (in Joules) is the area under a power-time curve. Calculate the energy stored in the inductor by estimating the area under your graph in the previous part.

h. Compare your answer in g to the value E = ½LI²_Max, where I_Max is the maximum value of the current that you found in part e.