# Find the voltage drop of each resistor?

A 6 ohm resistor is connected in parallel with a 4 ohm resistor. The parallel connection is placed in series with a 2 ohm resistor, and the entire circuit is placed across a 10 volt difference potential.

Relevance

6 ohm in parallel with 4 ohm is

Rp = (6 * 4) / (6 + 4) = 24 / 10 = 2.4 ohms

Rt = 2.4 ohms + 2 ohms = 4.4 ohms

Using voltage dividers:

http://en.wikipedia.org/wiki/Voltage_divider

V of 2.4 parallel resistance = 10 * 2.4 / (2.4 + 2) = 5.4545 Volts

which is the same across the 6 ohm and 4 ohm resistors.

V of 2 ohms = 10 * 2 / (2.4 + 2) = 4.5454 Volts across the 2 ohm resistor.

Sum of all voltage drops = 5.4545 Volts + 4.5454 Volts = 10 Volts

Source(s): ALOT of engineering experience.
• (6) ll (4Ohm) = 2.4 Ohm

V of 2 Ohm = [(20/10) / (44/10)]*(10V) = (200/440)*(10V) = (4 and 6/11) Volts

Since resistors connected in parallel must have the same Voltage

V of 4 Ohm = V of 6 Ohm = [(24/10) / (44/10)]*(10V) = (5 and 5/11) Volts

Proof:

V of [(6 Ohm) ll (4 Ohm)] + V of 2 Ohm = V total

(5 and 5/11)V + ( 4 and 6/11)V = 10 Volts

• hi why now not be conscious loop equations or Nodal equations on it. it is very individual-friendly that way. attempt assuming 2 loops, assume the present i1 flows interior the 1st and i2 interior the 2d. Write down Kirchoffs Voltage Equations for the two the loops. You ll get 2 equations on the subject rely of i1 and i2, freshen up them concurrently and likewise you have the easily values of i1 and i2, multiply them by way of making use of their respective resistances and likewise you ll get the voltage drops. keep in mind that for the time of r3, the sum or huge distinction of i1 and i2 pass, based upon the direction of i1 and i2 which you assume.

• The parallel combination has a resistance of 2.4 ohms, so the total resistance is 4.4 ohms and the current is 2.27 amps.