posted by christina on .
If the resistances R1= 16. ohms, R2 = 38. ohms, and R3 = 18. ohms, and the voltage V = 6.0 volts, what is the current ix (in Amperes) in the circuit.
tinypic .cm /r / nfr48k/ 8
The correct answer is .1443 but I want to know how they got that. Thanks.
The equivalent resistance from R2 and R3 is
1/R = 1/38 + 1/18
R = 12.214
So, we now have a total R of 16+12.214 = 28.214 ohms
That means our current through R1 and the R2R3 network is 6/28.214 = .212 amps
Now, divide that current between R2 and R3 and
Ix = .212(38/56) = .144
or so. Check my math and adjust sig figs.