Question

The drawing shows three different resistors in two different circuits. The battery has a voltage of V = 20 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.

Answer 1

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. 20/85=0.270A.

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= 11.76V is cut out at the first resistor.

V2 is 25/85(20) = 5.88V,

and V3 is 2.35V (add them all and you're close enough to 20 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 = 6.25?.

Now, the voltage stays the same, but the current drops. Remember V/R=I so, the voltage across each resistor is 20 and the first resistor is 50? = 0.4A,

A

At R2 it is 20/25 =0.80A, and

the last one is 20/10 for 2.0A Add all the amps to get 3.2A.