# Voltage divider with battery load

I'm trying to find the current across all the components but am having trouble doing so because I don't know how to deal with the extra battery in this case.

I've found that without the battery V(out)should be 4V, so is this cancelled by the battery? How does the battery add to current in this case?

• Where is Vout? (need more characters...) – Peter Bennett Apr 7 '17 at 6:31
• glad you figured it out in this hypothetical situation – Tony Stewart Sunnyskyguy EE75 Apr 7 '17 at 6:36

Since the battery will hold the voltage across the 4K resistor at 4 volts, and the voltage divider will also make the voltage across the 4K resistor 4 Volts, no current will flow to or from the battery.

All nodes voltages are determined by the batteries.

Simply write down what the node voltages are, and use Ohms Law to get the resistor currents.

The battery currents will be determined by the resistor currents.

Wiki says:

Superposition Theorem: The total current in any part of a linear circuit equals the algebraic sum of the currents produced by each source separately.

Since you don't have any current source just replace once at a time one voltage source with a short circuit, and leave the other intact. For example, if 10V is shorted first then the voltage across R2 would be 4V, hence a current of 4V/4koms = 1mA would flow through the resistor R2. Now assume 4V is shorted and 10V is on, then R2 would be shorted too and no current would flow through it. Hence the total current flowing through R2 becomes Itot = 1mA + 0A = 1mA.