Electrical Engineering - Series-Parallel Circuits - Discussion

Voltage drop is directly related to resistance.

Higher the resistance higher will be the voltage drop.

Voltage drops across internal resistances of the source, across conductors, across contacts, and across connectors are undesired; as supplied energy is lost (i.e. dissipated).

Voltage drops across loads are desired; as supplied energy performs useful work.

Hence in the above question the supplied energy will be put to maximum use when the value of load resistance is maximum.

As the question is asked for load resistance the answer will be 1M ohm.

If the question would have been for internal resistance then the answer would have been 330 ohm.

The question is pertaining to Voltage Divider, hence: V(output) = V(input) * (R1 / (R1 + R2))
Since the question is the LEAST effect of the Output Voltage we can then assume any value for V(input) in the example below I used 5 Volts, R1 or R (load) is connected parallel to the 3 series resistor because that is how voltage divider works and example would be like this.

V (out) = 5V * (1M ohms / (3K ohms + 1M ohms) = the answer would be appromately equal to 5, since the output and input are almost of equal value then R (load) which is 1 M ohms has the LEAST effect on the circuit.

Here voltage divider having the low values of resistors. These can allow the some what of fluctuations when the input voltage is get changes by means when we increase input voltage we will get effected output voltage but we are using high range resistor we get least effected output voltage. That means high range resistor strongly opposes the changes in voltage. Here the change will be the effect.

R1= 3(1k ohm) = 3k ohm.

Vout = (R2/(R1+R2))Vin.

Lets take minimum resistor R2 = 330 ohm.

Vout = (R2/(R1+R2))Vin.
Vout = (330/3330)Vin.
Vout = 0.099 Vin.

Lets take maximum resistor R2 = 1M ohm.

Vout = (R2/(R1+R2))Vin.
Vout = (1M/1003000)Vin.
Vout = 0.997 Vin.

Here you can observe, bigger the value of R2, smaller the effect on Vout. So, the answer is 1M ohm.

As the question is that among the options which has least effect on output voltage. As all option except 1M is less than or equal to 1K so the output of voltage divider has the effect on other 3 loads except 1M because for a high value load resistance the current for specified voltage will be low as compared to low load. So 1M will be effected least.

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 1k ohm then the result will be V (out) = V (In) (0.25).
So, the output voltage will be less affected if the resistance value will be more so the correct answer is 1M ohm.

First of all least effect is not big difference in voltage. I equals V/R resistance increases current decreases if we are using 330 as answer it will get big current and voltage drop will be big but question is least effect. So biggest value is answer so current will be less so voltage drop will be less.

Here we got R2 = 3kohm = 3000ohm.

Lets take minimum resistor R1 = 1kohm = 1000ohm.

Vout = (R2/R1+R2) Vin.

= (3000/1000+3000)Vin.

= 0.75Vin.

So Vout/Vin = 0.75.

Now for max resistor R1 = 1Mohm = 1000000ohm.

Vout = (R2/R1+R2)Vin.

= (3000/1000000+3000)Vin.

= 0.003Vin.

So Vout/Vin = 0.003.

As per Loss formulae: I^2.R & Ohm's Law: I inversely proportional to R.

More Resistance higher will be the loss.

1Mohm is the huge value of resistance.

Definitely, the output voltage is very less.

Hence, the answer for least effect is 300 ohm.

Otherwise, Data is insufficient.

As we connect one of the series resistance with load in parallel and we know that parallel of two things results in lesser than lesser thing, but larger the value of load resistance parallel will be nearer to lower resistance. So will affect v less so see largest resistance.