### Discussion :: Series Circuits - General Questions (Q.No.5)

Jeyaraj N said: (Feb 24, 2011) | |

ANS: A Just think when resistance keep on incresing accros the voltage source, there will not be a current at the maximum resistance, so ther will not be a power dessipation at max resistance (consider open). Imagine when your reistance is minimum (consider short) accros voltage source, current will be high and your power desipation also will be high. |

Lavanya said: (Jun 19, 2011) | |

Power= v*i =v* (v/r) = v^2 /r. P= v^2/r it implies that the resistance and power are having inverse relation, so by this we can conclude that as resistance is less power dissipation will be more. From given options one 220 ohm resistor is less compared to other options. |

Harpreet said: (Jul 7, 2011) | |

V=IR P=VI P=V^2/R WHEN R=220 then power will maximum now r is denominator so if we increses r then "P" will reduces so p will maximun when r=220 now in option anser is "A" one resistor in series so right answer in second option two resistor in series means R increses |

Anand said: (Jul 25, 2011) | |

Nice explanation Lavanya |

Shikha said: (Aug 9, 2011) | |

Very good explanation Harpreet. |

Balaji said: (Oct 7, 2011) | |

Nice explanation by lavanya. |

Hemraj said: (Dec 12, 2011) | |

In this case P=i^2*r power dissipation is perportional to the ressitace. So more resistace causes of more power dissipation. |

Sadik Pasha said: (Jan 4, 2012) | |

In series, number of resistors increase total resistance increase so resistance increase current decrease so power dessipation less. |

Himanish said: (Jan 30, 2012) | |

YES I am also having the same confusion.... P=VI V=IR So P = (I^2)R BUT again P = (V^2)/R These statements are contradictory but both are true... |

Mohit said: (Jan 30, 2012) | |

If you know that in domestic wiring we have parallel connection to minimize power dissipation. If we do in series then voltage will be low and power dissipation will be high. Think about it. |

Satishgnd said: (Mar 21, 2012) | |

Very good explanation Harpreet. |

Amarender said: (Jun 28, 2012) | |

In the series if we increase resistance one by one current will be lower than previous. Hence we know the P=V*I. Or in other words if we increase the resistance power dissipation will be low according to V^2/R. |

Ashu said: (Aug 25, 2012) | |

As we know current passes maximum through minimum resistive path. So as per the formula of power P=I^2*R. Square of I decreases with increase in resistance. So minimum the resistance maximum the power. |

Raviraj said: (Oct 5, 2012) | |

@Hemraj. For P=V^2/R In this case the resistance and power are having inverse relation, so by this we can conclude that as resistance is less power dissipation will be more. And for P=I^2.R here the resistance and power are having direct relation, So more resistance causes of more power dissipation. But as we know current through the CKT depends on Voltage, so we can take voltage as constant so as per P=V^2/R the resistance is less power dissipation will be more. |

Binay said: (Oct 21, 2013) | |

As p = v^2/r, here v constant, So pr = constant. So for max p, r should minimum. |

Krish said: (Jan 3, 2014) | |

Resistor is a device which resists the flow of power through it, The answer will be chosen from the least value of the options simple. |

Vara Prasad said: (Jan 20, 2014) | |

Power consumption formula is V^2/R. Here V is constant so whenever R value is low V^2/R value is more. So R value is low at single 220 ohms resistor. |

Dilip said: (Feb 8, 2014) | |

Power = v*2/r. If r is minimum then power has maximum value. And we know that i = v/r. And p = i^2*r directionally proportional. |

Ponjothirajalingam said: (Aug 17, 2014) | |

But at the same time Power is directly proportional to resistance only. |

Sahi said: (Jan 5, 2015) | |

Higher the resistance lower the power dessipation. |

Rupesh said: (Apr 26, 2015) | |

What about maximum power transfer theorem? Which says the max power will be transferred to the load when load impedance is equal to the source impedance. It means if load independence is greater than the source impedance then the power transferred to the load is less and vice versa. If we match both the incidences we can get max power and max heat dissipation. So the question is what is the value of source impedance of 120 volt source? |

Sriksnth said: (Apr 28, 2015) | |

P = v^2/r. Power is inversely proportionally to resistance. So, as resistance is less power will be more. Option A is correct. |

Ramesh said: (Jun 4, 2015) | |

Can any one explain? What happens if parallel combination is there? |

Rajesh said: (Jul 28, 2015) | |

For a single resistor voltage drop is maximum across that resistor, so power is maximum. But when two or more resistor come into picture then voltage drop across the resistance will be lesser than previous one so power will be less. |

Sanoop said: (Aug 19, 2015) | |

P = V^2/R. If R decreases then Power increases. P = I^2.R. If R decreases then current increases since I^2 is considered for power calculation, increase in current will be considered over increase in resistance. |

Satyaranjan Acharya said: (Sep 20, 2015) | |

Very simple thing friends P = V*I. In this case we can increase power by two process either by increasing current or by increasing voltage though power is directly proportional to the current & voltage. But in this question voltage is constant so we have to consider the current for solve this question. And current is higher in one resistance in comparision to series combination of 4 resistance. So the answer is "A". |

Apiaps said: (Oct 30, 2015) | |

v = IR = I = V/R as are reduces I increases. And P = I^2 *R. As R reduces I increases as p increases. |

Shamik said: (Jan 21, 2016) | |

P = V^2/R. As R increases, P decreases. |

Ramesh said: (Oct 12, 2016) | |

Nice explination @Sadik Pasha. |

Jayaram said: (Nov 24, 2016) | |

Nice explanation, Thank you all. |

Susanta said: (Dec 15, 2016) | |

When power are series it calculated as parallel so, many numbers resistor has the power are too low then the single 1/p1 + 1/p2 + 1/p3 ...< p1. |

Vijay said: (Dec 30, 2016) | |

Nice @Lavanya. |

Angesh said: (Apr 16, 2017) | |

Why p=i^2*r is not considered? |

Shambhavi said: (Jun 30, 2017) | |

P=i^2 R i.e., P=V^2/R. So if the resistance goes on decreasing, the corresponding power is less. So the minimum resistance i.e.,220 ohm delivers more power. |

Syed said: (Aug 20, 2017) | |

Clear explanation, thank you @Satyaranjan Acharya. |

Ajith Kumar A said: (Sep 14, 2017) | |

P is inversely proportion to R. |

Jay Chandra said: (Jan 20, 2018) | |

Nice @Lavanya. |

Amarnath Maity said: (Feb 5, 2018) | |

P = V^2/R. Choose the lowest total resistance to maximize the dissipation. |

Lio said: (Aug 28, 2018) | |

In series, current will remain the same hence; P=i^2r. Hence resistance directly proportional to power dissipation. Then, How option A is correct? |

Abdul Majid said: (Dec 13, 2018) | |

@ALL. Power, P has two formulas: #1>> P=V^2/R and #2>>P=I^2*R The first relation is valid when Voltage is constant for all resistors (i.e., for Parallel combination of Rs and V. 2nd Relation is for the series circuit when Current, I is constant for all resistors. In this question, 2nd relation is being used, Now, by increasing Resistance, P increases, Its true But. Value of current decreases " Square times".Hence, the Overall value of Power dissipation decrease with increase in resistance for series circuits. I hope this will help. |

Mayank said: (Sep 11, 2019) | |

For an Electrical circuit. P=I*I*Req, Req= equivalent resistance of the circuit. We can also write it as P=V*V/Req. Now case1- one 220ohm resistance Req=220ohm, case2-Two 220ohm resistance. Req=2*220ohm ,P2=P1/2, Similarly P3=P1/3. P4=P1/4. Hence max power is dissipated in the case 1. |

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