Yes, wiring resistor in parallel lowers the resistance. The formula is 1/((1/R1)+(1/R2)+ ... (1/Rn))
So the net resistance of a 300 and 400 ohm resistors in parallel is 1/((1/300)+(1/400)) = 171 ohm
I don't know what you mean by 'putting out 4V'. There is a voltage drop accross a resistor which is related to the resistance and current by the formula V=I x R (Voltage = Current x Resistance)
Let's take example 1.
- circuit1.gif (3.2 KiB) Viewed 2675 times
If the battery is producing 5v, you have a 400 ohm resistor and a voltage drop accross the resistor of 4v, this means the voltage drop across the load is 1v. We need to know what is the current in the circuit and what is the resistance of the load. Then we can work out what will happen when we add the second resistor.
V = i x R
4= i x 400
i = 4/400 = 1/100 = 0.01 amp
So the current in the circuit is 0.01 amps.
Calculate the resistance of the load
V=i x R
1=1/100 x R
R= 100
Now we add the second resistor in parallel, the total resistance of the circuit is now 171 +100 = 271 ohms.
- circuit2.gif (3.91 KiB) Viewed 2675 times
V=i x R,
5 = i x 271
i = 5/271
i = 0.01845
The voltage drop across the two resistors is
V = i x R
V = 0.01845 x 171 = 3.16 volts
And the voltage drop across the 100 ohm load
V = 0.01845 x 100 = 1.84
which could also be calculated by subtracting 3.16 from 5v
Hope that helps ...