Question
Download Solution PDFA wattmeter is measuring the power supplied to a circuit whose power factor is 0.7. The frequency of the supply is 50 cycles/second. The wattmeter has a potential coil circuit of resistance 1000Ω and inductance 0.5H. The error in the meter reading is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFPower factor, cos ϕ = 0.7
⇒ ϕ = cos-1(0.7) = 45.57°
⇒ tan ϕ = 1.02
Frequency (f) = 50 Hz
Potential coil resistance, (Rp) = 1000 Ω
Inductance (Lp) = 0.5 H
Xp = ωLp = 2πf Lp
= 2π × 50 × 0.5 = 157.08Ω
\(\tan \beta = \frac{{{X_p}}}{{{R_p}}} = \frac{{157.08}}{{1000}} = 0.157\)
Percentage error = ±tan ϕ tan β × 100
= ± 1.02 × 0.157 × 100
= ± 16.02%Last updated on Jul 2, 2025
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