GATE EE 2006 Electrical Circuits Q 2 - Grad Plus

# GATE EE 2006 Electrical Circuits Q 2

An energy meter connected to an immersion heater (resistive) operating on AC 230 V, 50HZ, AC single phase source reads 2.3 units(kWh)in 1 hour. The heater is removed form the supply and now connected to a 400V peak to peak square wave source of 150Hz. the power in kw dissipated by the heated will be

a)3.478

b)1.739

c)1.540

d)0.870

Ans. (b)

Explanation:

Let’s sar resistance of heater is R.

i) When heater connected to 230 V, 50Hz source energy consumed by the heater =2.3 units or 2.3 kWh in 1 hour

Power consumed by the heater is given as

P_1=\frac{energy}{time\;period}=\frac{2.3kWh}{1h}=2.3kW

P1 = 2.3 kW

RMS value of the input voltage is given as, Vrms = 230V

Hence, P_1\;=\frac{{V^2}_{rms}}R

2.3\times10^3\;=\;\frac{230^2}R R\;=\;23\Omega

ii) When heater connected to 400V (peak to peak ) square wave source of 150 Hz. It is shown below.

Vrms of this input voltage is given as,

V_{rms\;}\;=\left[\frac1T\int_0^TV^2\;dt\right]^\frac12 V_{rms}=\left[\frac1T\left(\int_0^{T/2}200^2\;dt\;+\int_{T/2}^T-200^2\;dt\;\;\right)\right]^{1/2}

Vrms = 200V

P_{2\;}\;=\;\frac{V_{rms}^2}R\;=\frac{200^2}{23}\;\times\;10^{-3}\;kW

=1.739 KW

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