a 12 volt series circuit has a 2,4,5 and 2 ohm resistor. the voltage drop across the last resistor is?

Find first the total resistances

R total = 2+4+5+2 = 23 ohms

Find the total current in the circuit

I total = V total / R total

= 12 V / 13 ohms

= 0.923 amp.

Current drop at 2, 4, 5, 2 is equal total current which is 0.923

Voltage drop at the last resistance which is 2 ohms

V @ 2 ohms = I R = 0.923(2)

V = 1.846 approximately 2 volts

The answer list provided does not contain the right answer. As others above have noted, the correct voltage is 1.84 Volts.

The voltage across a particular resistor in a simple series circuit is proportional to that same resistor's part of the total resistance of the string. In this case, it is 2/13ths of the 12 volts applied to the string of four series resistors.

(2/13)12 = 1.8461538461538461538461538461538 Volts

Well, you get any voltage drop by multiplying the Resistance by the Current in this case you know the current is constant because the circuit resistors are series so then use you Ohms law, which is I= E/R OR Current= 12 volts / 13 ohms, I = 0.923A then figure the drop over that one resistor by E= I*R OR Voltage = 0.923A * 2 Ohms, Voltage = 1.846. And there you go.

I'm not sure I understand the question. What is a 2,4,5 resistor?

If the elements are 2, 4, 5 and 2 ohms and the source voltage is 12V, then the drop across the last resistor is calculated per the voltage divider rule, which tells us that the voltage across a series element is calculated as the resistance of that element divided by the sum of the resistances multiplied by the source voltage.

Thus we have 2ohms/(2+4+5+2)ohms*12V = 24/13V.

Since this is not one of your answers one of us is subject to a misunderstanding.

E = I x R loop entire loop has 12 volt drop, 13 ohms resistance

12 = I X 13

I = 12/13 entire loop has same current

E = I x R resistor

E = 12/13 x 2 = 24/13 volts

if there are a total of five resistors in series , 2,4,5, and 2 ohms that adds up to 13 ohms , divide that into 12 volts. gives .923 amps and times 2 ohm last resistor gives 1.84 volts across last resistor.

Since you are using a series RLC circuit, I will assume the power source is AC, therefore, the peak voltage across the resistor will equal the power source, so the I = 3.71 / 4.14 = 0.90A at Phase angle 0. If the voltage source is DC, the current will be 0, since the capacitor is an open circuit to a DC power source

There is no correct answer. Im guessing tha the question is mistyped. and one of the two ohm resistances is supposed to be one ohm. With that corrected, the correct answer is two volts. As written, the answer would be appx. 1.85V To solve, Kirchoffs law says the sum of voltage drops in a circuit is equal to the voltage applied to the circuit. so twelve volts is dropped by 13 ohms of resistance ot 12/13V per ohm. Two ohms would drop 24/13V or the 1.85V

2 volts

2 volts