Electrical Engineering - Energy and Power - Discussion
Discussion Forum : Energy and Power - General Questions (Q.No. 1)
1.
A 33
half-watt resistor and a 330
half-watt resistor are connected across a 12 V source. Which one(s) will overheat?


Discussion:
58 comments Page 5 of 6.
Presley C said:
8 years ago
33 Ω resistor will overheat as the connection is parallel it's not specific. So, the Answer is A.
Subhrajit jena said:
8 years ago
According to power loss formula, 330 ohm resistor will produce overheating because both are connected in series.
Vinny said:
7 years ago
In question, if both resistors are connected in series across 12v supply, then answer is D, either the A correct answer.
Akshay jadhav said:
7 years ago
Power of both are same so only current will be change. The output of the resistor in the form of heat that heat output is remains same for both for same wattage.
J D D said:
7 years ago
The resistor will overheat because the capacity of both resistors are 5watt And due to 12v supply no one will take higher current than it's rated capacity. (respectively i1=0. 3636, i2=. 03636).
Swetha said:
7 years ago
The Key point is;
Less resistance value resistor will dissipate more heat.
The connection is a parallel circuit.
Formulae is Power = V*V/R.
Less resistance value resistor will dissipate more heat.
The connection is a parallel circuit.
Formulae is Power = V*V/R.
Abdul Ghafoor said:
7 years ago
The heat is cause by current flowing through it so these two resistors are connected in series and current is same across both so neither resistor is overheated because value of current is very less.
Sher shah said:
7 years ago
Both the resistors have same power rating so no one will be heat up.
D Shwetha said:
7 years ago
Here, it is given that, the resistors are 0.5W & also a parallel connection.
Heat is directly proportional to power (H=P*t). So, power rating 0.5W & Heat is also both resistors are same.
Heat is directly proportional to power (H=P*t). So, power rating 0.5W & Heat is also both resistors are same.
Junaid Ahmed Hassan said:
7 years ago
By Ohm's law, V=I * R.
Power dissipated across the resistor, P=I^2 x R.
Let R1=33 ohm, R2=330 ohm.
So, let's calculate the current through each resistor. We know V = 12V.
Current through the resistor R1, I(R1)=V/R1= 12/33= 0.36 A.
Power dissipated by R1, P(R1)= I(R1)^2 x R1 = 0.36^2 x 33 = 4.27 Watts.
Current through resistor R2, I(R2)=V/R2=12/330=0.036 A.
Power dissipated by R2, P(R2)= I(R1)^2 x R2= 0.036^2 x 330 = .42 Watts.
Power dissipated across the resistor, P=I^2 x R.
Let R1=33 ohm, R2=330 ohm.
So, let's calculate the current through each resistor. We know V = 12V.
Current through the resistor R1, I(R1)=V/R1= 12/33= 0.36 A.
Power dissipated by R1, P(R1)= I(R1)^2 x R1 = 0.36^2 x 33 = 4.27 Watts.
Current through resistor R2, I(R2)=V/R2=12/330=0.036 A.
Power dissipated by R2, P(R2)= I(R1)^2 x R2= 0.036^2 x 330 = .42 Watts.
(5)
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