- 2070 Chaitra · 6 marks
Power ratio of two resistors at 70°C
Short answer
At 15°C resistances are equal. At 70°C, R1 = 1.220R, R2 = 1.275R. Since P ∝ 1/R, P2/P1 = 1.22/1.275 ≈ 0.957.
Two resistors made of different materials having temperature coefficients of resistance α1 = 0.004/°C and α2 = 0.005/°C are connected in parallel and consume equal power at 15°C. What is the rate of power consumed in resistance R2 to that in R1 at 70°C?
Resistors in parallel have the same voltage. Equal power at 15°C therefore means equal resistance at 15°C: P = V²/R.
Step 1: Resistances at 70°C
Let R₁ = R₂ = R at 15°C (α values are taken at 15°C).
R₁(70) = R [1 + 0.004 × (70 − 15)] = 1.220 R R₂(70) = R [1 + 0.005 × (70 − 15)] = 1.275 R
Step 2: Power ratio
At the same voltage, P ∝ 1/R.
P₂/P₁ = R₁/R₂ = 1.220/1.275 = 0.957
Answer: P₂ : P₁ ≈ 0.957 (R₂ consumes about 4.3 % less power than R₁)
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Series-parallel resistors, find fourth resistance · Basic Electrical Engineering, chapter 1
- Resistors A, B in parallel with C in series · Basic Electrical Engineering, chapter 1
- Delta connected coils line current and power · Basic Electrical Engineering, chapter 8
- Total power of series-parallel circuit · Basic Electrical Engineering, chapter 6
- Transformer coil temperature and power rating · Basic Electrical Engineering, chapter 1
- Series-parallel circuit currents and power · Basic Electrical Engineering, chapter 6
- IDMT grading: R1, R2 with 1000/5 CTs · Switchgear and Protection, chapter 7 · asked 2×
- Show i1 = i2 = -V/(R1+R2+R3) at t=0+ · Electrical Circuit Theory, chapter 2
- 3rd order Butterworth LPF, R1 = 1 Ω, R2 = 4 Ω · Filter Design, chapter 5 · asked 4×
- Significance of reflection coefficient; R1 = 1 Ω, R2 = 4 Ω · Filter Design, chapter 5