Electrical Engineering - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 5)
5.
Which of the following series combinations dissipates the most power when connected across a 120 V source?
one 220 
resistor
resistortwo 220 resistors
resistorsthree 220 resistors
resistorsfour 220 resistors
resistorsDiscussion:
44 comments Page 3 of 5.
Mayank said:
6 years ago
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.
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.
Sajmun said:
5 years ago
See.
When any resistor connected in series combination, then. Power is directly proportional to Current (I). So, A/q to find current _we have voltage and resistance.
Therefore, P=[I]2 (square)R. So, when we find the value of current, we will see that (a) option have a maximum value.
When any resistor connected in series combination, then. Power is directly proportional to Current (I). So, A/q to find current _we have voltage and resistance.
Therefore, P=[I]2 (square)R. So, when we find the value of current, we will see that (a) option have a maximum value.
Amarender said:
1 decade ago
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.
Or in other words if we increase the resistance power dissipation will be low according to V^2/R.
Lavanya said:
1 decade ago
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.
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:
1 decade ago
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
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:
1 decade ago
Nice explanation Lavanya
Shikha said:
1 decade ago
Very good explanation Harpreet.
Balaji said:
1 decade ago
Nice explanation by lavanya.
Hemraj said:
1 decade ago
In this case
P=i^2*r power dissipation is perportional to the ressitace. So more resistace causes of more power dissipation.
P=i^2*r power dissipation is perportional to the ressitace. So more resistace causes of more power dissipation.
Sadik pasha said:
1 decade ago
In series, number of resistors increase total resistance increase so resistance increase current decrease so power dessipation less.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers