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.
To calculate the current in the circuit, you can use Ohm's law, which states that the current (I) flowing through a conductor is equal to the voltage (V) applied across it divided by the resistance (R) of the conductor.
In this case, you have three resistances (R1, R2, and R3) connected in parallel. In a parallel circuit, the total resistance (R_total) is calculated as the reciprocal of the sum of the reciprocals of each resistance.
1/R_total = 1/R1 + 1/R2 + 1/R3
1/R_total = 1/16 + 1/38 + 1/18
To simplify this equation, we need to find a common denominator for the fractions:
1/R_total = (38*18 + 16*18 + 16*38)/(16*38*18)
1/R_total = (684 + 288 + 608)/(10944)
1/R_total = 1580/10944
Now we can find R_total by taking the reciprocal of both sides:
R_total = 10944/1580
R_total ≈ 6.93 ohms
Next, we can use Ohm's law to calculate the current (ix) in the circuit:
Ix = V/R_total
Ix = 6.0 / 6.93
Ix ≈ 0.865 Amperes
So the correct answer for the current (ix) in the circuit is approximately 0.865 Amperes, not 0.1443 Amperes as mentioned. Double-check the calculation or the source.