Ok. This will be kind of hard because there is a diagram and a table to accompan

Question

Ok.  This will be kind of hard because there is a diagram and a table to accompany this particular problem.  I need to have the work shown for this.

So the question is as follows... For the compound circuit, V=12V, R1=4 ohms, R2=6 ohms, and R3=3 ohms.  Show that the power supplied by the battery is equal to that dissipated in the resistors.  What principle does this illustrate?  Show work and use the accompanying table.  (for a DC circuit: Power= I^2 times resistance.

The table is set up as follows:  A top row of three headers which are "Circuit element", "Current", and "Power Dissipated".  Going down the column under "Circuit element" all values are given and are as follows... R1=4 ohms, R2=6, R3=3, and battery V=12.    The associated "Current" and "Power dissipated" cells which correspond horizontally to each R1, R2, R3, and battery V are blank.

The diagram of the circuit is a rectangle which is laid out lengthwise.  The battery is coming in from the center of the left vertical side.  At the top of the rectangle on the left side is resistor R1.  Moving to the right we come to an intersection.  There is a vertical line bisecting the rectangle and in the middle of this line is R2.  R3 is at the far right of the rectangle in the middle of the vertical side.  So we have a series/parallel circuit with R2 and R3 being the parallel.

I hope this paints a relatively clear picture for anyone who is nice enough to answer this!  Thank you!

Explanation / Answer

Find Help at http://academicresearchpapers.org/ if stuck or confused

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.