T The drawing shows three different resistors in two different circuits. The bat
ID: 2274570 • Letter: T
Question
T
The drawing shows three different resistors in two different circuits. The battery has a voltage of V = 26 V, and the resistors have resistances of R1 = 50.0 ?, R2 = 25.0 ? and R3 = 10.0 ?. Determine the current through and the voltage across each resistor.
(a)
(b)
Explanation / Answer
Always start with V=IR.
In a series circuit,you add the resistance values of each resistor to get the total resistance. So 85 Ohms. Now find the total current. 23/85=.270A. In a series circuit, the current is constant. So there are your first 3 answers. The Voltage drops proportionally across each resistor. 50 ohms is 50/85 of the total, or .5882 multiply the voltage by that to see the drop. 13.529V is cut out at the first resistor. V2 is 25/85(23) or 6.765V, and V3 is 2.705V (add them all and you're close enough to 23 total volts.)
The next circuit is parallel. Take the inverse of each resistor (1/R) add them all together, and take the inverse of that answer to find the total resistance. (1/R1 + 1/R2 + 1/R3)^-1 to get .16^-1 or 6.25?. Now, the voltage stays the same, but the current drops. Remember V/R=I so, the voltage across each resistor is 23 and the first resistor is 50? or .46A, at R2 it is 23/25 or .92A, and the last one is 23/10 for 2.3A Add all the amps to get 3.68A. Check it with 23/6.25 and your done
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