Electrical Engineering - Series-Parallel Circuits - Discussion

The load resistor is connected across any one 12k resistor.

Now consider 1M resistor is connected across 12k then total resistance become 8K.
Now consider 24K resistor is connected across 12k then total resistance become 11.85k.
Now consider 18k resistor is connected across 12k then total resistance become 7.2k.
Now consider 12k resistor is connected across 12k then total resistance become 6k.

So out of 8k, 11.85K, 7.2K and 6K, the voltage across 6k is small.

So this resistor reduces the output voltage. So 12k resistor have the more effect on output voltage.

If we connect 12k load resistor then output voltage reduces to low value, compared to other load resistance value.

Now consider 1M resistor is connected across 12k then total resistance become 11.85K.

Now consider 24K resistor is connected across 12k then total resistance become 8K.

Now consider 18k resistor is connected across 12k then total resistance become 7.2k.

Now consider 12k resistor is connected across 12k then total resistance become 6k.

So out of 8k, 11.85K, 7.2K and 6K, the voltage across 6k is small.

So this resistor reduces the output voltage. So 12k resistor have the more effect on output voltage.

The correct answer is D.

The answer cannot be A or C, because they given that two 12k resistors are in series, so 1M and 18K doesn't come into picture.

Now the answer may be 12K or 24K.

A voltage divider circuit must contain at-least two loads and we should take output across any one load. If we take voltage across total load then it is not called a voltage divider circuit.

So in this problem we should take voltage across any 1 12K resistor. Hence 12k resistor only can have effect on output voltage.

@All.

First understand the question.

For voltage divider circuit, V_out=V_in×(R2/(R1+R2)); R2-Load resistance to be changed. If we assume V_in=50V.
For R2=1Mohm, V_out~49V.
For R2=18Kohm, V_out=30V. For R2=12Kohm, V_out=25V.

That's why, 12Kohm resistance has the most reduction in output voltage.

So it affects more. I hope you will understand.

The formula for output voltage is;

V (out) = V (In) [ (RL / (R1 + RL) ]
So, if we take 1M ohm then the result will be V (out) = V (In) (0.9).
And if we take 12k ohm then the result will be V (out) = V (In) (0.33).
So, the output voltage will be more affected if the resistance value will be less so the correct answer is 12k ohm.

In voltage divider we take parallel of one of series resistance with load resistance. Now if we take load resistance equal than parallel of two will be half of one of the series resistance which will be minimum and hence affect the voltage more.

We know that from ohm's law if current increases voltage also increases. So if resistance decreases current increases. 12k+12k gives 24k.

In parallel resistance decreases. So 12k is more appropriate for less resistance. So more voltage.

By ohm's law V=IR
I=V/R

SO CURRENT INVERSELY PROPORTIONAL TO THE RESISTANCE

Hence less will be resistance more is the current flowing so more voltagee dropped and low resistance will more effect on output voltages.

So answer is D

BY OHM LAW.

Inversally proportional to the resistance so less will be the resistance more will be the current hence less valued resistane effectt more voltage output.

SO ANSWER IS D=12.

Inversely proportional to the resistance so less will be the resistance more will be the current hence less valued resistance effect more voltage output.

By ohm's law V = IR.
I = V/R.