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.

1 answer

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.