Electrical Engineering - Series-Parallel Circuits - Discussion

Assume a voltage source of 5V.
R1 = 10k ohms.
R(load) = 10k ohms.
Note that the load resistor is connected parallel to R1.

Rtotal = (10k*10k)/(10k + 10k) = 5k ohms,
I = 5V/5k ohms = 1mA

Removing Rload
I = 5V/10k ohms = 0.5mA.
Thus, indicating that removing Rload will results in decreasing the current from the source.

Load is connected in parallel to the second resistor in the voltage divider. If you connect two resistors in parallel, it is inevitable that the resulting resistance is lower than any of the two resistors in parallel. Therefore, by removing the load, the total resistance is the resistance of the remaining resistor from the voltage divider, and it is greater than the resistance when it is parallel with the load.

Since the resistance became larger, the current drawn decreased.

You can also think of it as the load consuming more current, and therefore when you remove the load, there is less consumption of current.

In an voltage divider circuit, for no-load, it is called an open circuit hence current decreases.

This is because in the voltage divider circuit consider there are two resistors one top resistor is connected to +vcc and the other ends of the top resistor is connected to the second resistor and output is taken across the second resistor. The junction of 1st and second resistor is grounded if it is not grounded, according to the question-answer is D.

If the grounded answer is A.

I'm agree @Noxa.

My explanation is; If load removes means current increase.

2 Scenarios:

First, if the load resistance is removed from the circuit, then it could create an open circuit which would mean that the current flowing = 0 (cutoff).

Second, in case the load resistance is lowered in a series circuit, then the supply current would increase due to ohm's law.

I don't know why the answer is 'decrease', Please explain.

@Nivi.

Because the circuit is not open-circuited.

Maybe, The circuit is short-circuited.

@Abhishek.

In voltage division rule,
The resistors are in series not parallel.
So, in series voltage is same and when the resistor is removed it acts as O.C CRT.
So, voltage is high and current is low, ans :A.


Cut off means there should be some limited minimum resistance value, as a resistor is O.C CRT no minimum resistance, so cut off is wrong.

In question, it is given voltage divider circuit. In voltage divider circuit, we have a resistance in parallel to load.

If the load is removed, effective resistance increases and current decreases.

IN SC "I" high, "V" low.

IN OC "I" low, "V" high. It's inverse notation.