What is the net resistance of the circuit connected to the battery in Fig. 19-41? Each resistance has R = 1.8 k.
Click on the link below. It will bring up the figure of 19-41. There are 12 V in this problem.
http://i24.photobucket.com/albums/c39/immeteen/19_33.gif
2007-09-19 10:49:06 · 4 answers · asked by Jessie L 2 in Science & Mathematics ➔ Physics
Let's take it in parts.
From A to B you have two parallel resistances, 2R and R
Net = (1/2R + 1/R)^-1
= 2R/3
This is in series with R (5R/3) and both are in parallel with another R.
Net = (3/5R + 1/R)^-1
= 5R/8
Now this is in series with another R.
Net = 13R/8 = 2.925 k-ohms
2007-09-19 10:56:28 · answer #1 · answered by Dr D 7 · 2⤊ 0⤋
The net resistance is (13/8)R.
2007-09-19 10:59:49 · answer #2 · answered by supastremph 6 · 0⤊ 0⤋
think of of it this way, as in case you had resistors (each and all the comparable ohm fee) to paintings with fairly of cord strands. in case you parallel them, as strands could make a cable, the resistance is dwindled via the style of resistors or strands used. only divide resistance. If related end to end (or in series) the resistance retains increasing via the style of resistors used. only multiply resistance.
2016-12-17 05:25:25 · answer #3 · answered by Anonymous · 0⤊ 0⤋
1/2r+2/2r=3/2r,,,r1=2r/3+1=5r/3
1/r2=1/r +3/5r =8/5r,,,,,,,r2=5r/8
r3=r2+r
r3=5r/8 +8r/8=
r3=13r/8=13(1.8)/8=~2.7 k ohm
I= V/r3 = 12 volt /2.7 000 =~4.8 mV
2007-09-19 11:04:14 · answer #4 · answered by Anonymous · 0⤊ 1⤋