Ratio of thermal energy released in two resistor R and 3R connected in parallel in an electric circuit is :
Question
Solution
The ratio of thermal energy released in two resistors R and 3R connected in parallel in an electric circuit can be found using the formula for power dissipated in a resistor, which is P = V^2/R, where V is the voltage across the resistor and R is the resistance.
Step 1: Since the resistors are connected in parallel, the voltage across each resistor is the same. Let's denote this common voltage as V.
Step 2: The power dissipated in resistor R is P1 = V^2/R.
Step 3: The power dissipated in resistor 3R is P2 = V^2/(3R).
Step 4: The ratio of the thermal energy released in the two resistors is P1/P2 = (V^2/R) / (V^2/(3R)) = 3.
So, the ratio of thermal energy released in the two resistors R and 3R connected in parallel is 3:1.
Similar Questions
Two resistors R1 and R2 of 4Ω and 6Ω are connected in parallel across a battery. The ratio of power dissipated in them, P1: P2 will be.
Using Ohm’s law, state three expressions for the electrical power that is dissipated in a resistor
Three parallel resistors have a total conductance of 2 mS. If two of the resistances are 1 and 5 kR,what isthe third resistance?
The power dissipated in a conductor of resistivity ρ connected across a battery is directly proportional to Only one correct answerA.ρ2B.√ρρC.ρD.ρ–1
Describe the difference between resistors connected in series and resistors connected in parallel.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.