# Homework Help: Voltage across a load resistor in a Voltage Divider Circuit

1. Jun 1, 2012

### Martinet

1. The problem statement, all variables and given/known data

Show that V across Resistor Rkohms is: (11R) / (3.96 + 4R) V

The circuit is a 5V cell in series with a 1.8k resistor. Following the 1.8k resistor are a 2.2k and Rk resistor in parallel with eachother. The wires then rejoin and return to the cell.

Current through the circuit has already been found as 5(2.2 + R) / (3.96 + 4R) mA

2. Relevant equations

I = V/R

3. The attempt at a solution

I = 5(2.2 + R) / (3.96 + 4R)
thus I through R
= 5(2.2 + R) / (3.96 + 4R) * (R / 2.2 + R) <= Current is divided proportionally to the strength of the resistor.
= 5R / (3.96 + 4R) <= Simplification.

V = IR
thus V across R
= (5R / 3.96 + 4R)) * R <= Voltage is equal to the divided current * the resistance, R.
= 5R^2 / (3.96 + 4R) <= Simplification.

I'm not quite sure where I'm going wrong. I have a funny feeling it might have something to do with voltage divider theory but I can't see how that would be incorporated. ANy help would be much appreciated.

2. Jun 1, 2012

### Staff: Mentor

Hi Martinet! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]
Correct.
Wrong. And it's still wrong even when you add the essential brackets.

https://www.physicsforums.com/images/icons/icon2.gif [Broken] Current in R = 5(2.2 + R) / (3.96 + 4R) * (2.2 /( 2.2 + R)) mA

Last edited by a moderator: May 6, 2017
3. Jun 1, 2012

### Martinet

Hey Nascent. Can you explain to me why the current is relative to the 2.2k as opposed to the resistor across which voltage is being measured?

4. Jun 1, 2012

### Staff: Mentor

The greater the parallel resistor, the greater the share through R. Current share is in inverse proportion to the resistances.

If the parallel resistor were zero ohms (a short circuit), then zero current would go through R (it would all be going through the short circuit).

5. Jun 1, 2012

### Staff: Mentor

With the pair being in parallel, measuring the voltage across one is the same as measuring the voltage across the other.

6. Jun 1, 2012

### Martinet

Okay. Now I understand. Vaguely, but I think I understand. Thanks very much. I think I need to refine my basic understanding of electricity itself. I'm still in that rut where I imagine electrons coming out of a battery instead of merely being pushed/dragged along by it.

7. Jun 1, 2012

### Staff: Mentor

It's easy enough to work out from first principles. Imagine two resistors in parallel, R and 2.2Ω, with a voltage V applied across each. The current through the 2.2Ω is V/2.2, the current through R is V/R.

The sum of the currents = ....
The fraction of total current that goes through R = (V/R) / ( ....... )

Reduce this expression to a simple fraction.

Commit the result to memory.