how do i solve series-parallel circuits with an open open circuit?

Question

you can find the problem here: http://www.geocities.com/shinryuken18/assignent.jpg

i am trying to find out how to solve for R6 or if the resistor is even to be included in the problem. i'm trying to find the voltage and current through each resistor

Answer 1

Let's think logically. Voltage drop ONLY ocurrs when there is a current flowing through the resistance..... (keep this in mind)

To solve this circuit, first, notice, the loop where the current will flow is consisting of R1, R7, and the series/parallel circuit consisting of R2, R3, R4, and R5. So, let's figure out the current value through this loop first.

Once you have it, you can figure out the voltage drop at R1. Then subtract that from 10 volts. You will then have the voltage at where R6 and R4 connects in reference to the ground.

Since R6 has no current flowing through it, the voltage drop is zero. Hense, the voltage at Vc in reference to ground is the same as the junction of R6 and R4.

I am assuming, you know how to calculate series and parallel resistance.

To solve for current and voltage, keep in mind, the voltage drop is current * resistance. When the current has to branch, incoming branch to a point equals outflowing current. Figure out the total resistance of the 4 resisters to come to the voltage differential first. Then take two series pair each and their resistance value to figure out the current flowing into it. Then solve for voltage drop individually for each resister.

(this is nearly impossible to explain in text only)

Answer 2

R6 will not affect Vc. Vc will be the same on both sides of R6. The assumption (in this case correct assumption) is that any voltage measuring device is an open circuit, so no current will flow through R6, therefore no voltage will exist across R6.

If you still need help with the rest of the circuit, let us know.

.