Electrical Engineering - Series-Parallel Circuits - Discussion

Assuming that from a voltage divider circuit with two resistors given (R1 and R2), we added a load resistance RL parallel to R2, making it a simple series-parallel circuit.

Note that (R2 parallel to RL) will decrease the total resistance.

1. Decreasing the total resistance will increase the total current drawn from the source: IT = 'VT/RT".
2. If the current drawn from the source increase, the current across R1 will also increase.
3. Consequently, the voltage drop across R1 will also increase, leaving less voltage for the output.
4. Therefore, the readings across Vo will decrease.

@Spurgeon: voltage divider is basically a circuit in which output voltage is fraction of the supply voltage. Actually in the secondary side of transformer there is voltage divider. So when decider moves number of turns that is impedance changes and accordingly we get different voltages in the load side. If there will be two impedance in series and we use V.D between the impedance then output voltages is.

V(output) = V(input) *Z2/Z1+Z2.

Let consider in this divider, Rx is the resistance of that resistor. Now we connect some load (its resistance may less or more than Rx) since load and divider resistance is in parallel system resistance will decrease. The current will increase and the voltage drop will in accordance with current. Conclusion voltage will decrease across that Rx.

Hello, everyone, I faced an issue with voltage drop.

I am using 24v supply source and connecting few load in parallel with it but when I am connecting one of their load voltage decreases by 10v at the source side and when I am removing this load from circuit voltage increases to 24v.

Please help me to solve my issue.

Voltage divider rule is applicable to only series connected elements. In parallel circuit even though if you increase load are decrease voltage remains same.

In this ex your connecting the load (nothing but resistance are inductance) definitely there would be drop of voltage across this load elements that's it.

When you connect the load,it means you have completed the circuit for current flow through it. Now the; Output voltage = Supply EMF (voltage at no load) - voltage drop (across the internal resistance of the source/supply).

That's why the output terminal voltage decrease in the case of a load. Option A.

It's all about series parallel circuit, generally voltage devices when there is series circuit and in parallel circuit voltage remain same.

So here load is connected it means another series load added means voltage must decrease.

When there is no load there is no current hence the internal resistance drop is zero...but when we connect a load current starts flowing. Hence some vol drop will appear across internal resistance.

Hence from KVL V-Ir=Vo.

Because when we apply load (resistance) then we create hindrance against the flow of electrons and hence voltage decreases like the example of bottle when we make a hole in bottle then level of water decreases gradually.

The output of certain divider is 12v that means the divider connected in parallel voltage remains same when load was connected it became series with divider hence voltage changes that is it decreases.