JEE MainPhysicsCurrent Electricity
In a circuit, a battery is connected to a resistor R₁ = 2 in series with a parallel combination of two resistors R₂ = 3 and R₃ = 6 . If the electrical power dissipated in the resistor R₂ is 12 W , the power dissipated in the resistor R₁ is:
Options
- A8 W
- B12 W
- C18 W
- D32 W
Correct answer
C. 18 W
Step-by-step solution
The power dissipated in R₂ is given by P₂ = I₂^2 R₂ . 12 = I₂^2 3 I₂^2 = 4 I₂ = 2 A The voltage across the parallel combination of R₂ and R₃ is: V_p = I₂ R₂ = 2 3 = 6 V The current through R₃ is: I₃ = V_p R₃ = 6 6 = 1 A The total current in the circuit, which flows through R₁ , is: I_ total = I₂ + I₃ = 2 + 1 = 3 A The power dissipated in R₁ is: P₁ = I_ total ^2 R₁ = (3)^2 2 = 18 W Answer: 18 W