Name: Series RL Circuit in AC Supply
Uploaded: 2019-07-19
Duration: 7 min 29 s
Description: Let's explore the current equation, power equation, and power plot for a series RL circuit with an AC supply by watching this informative video. Watch and le...

Question

The maximum values of the alternating voltages and current are

400 V and 20 A

, respectively, in a circuit connected to

50 Hz

supply and these quantities are sinusoidal. The instantaneous values of the voltage and current are

200 \sqrt{2} V and 10 A

, respectively. At that instant both are increasing positively. Determine the average power consumed in the circuit.

Accepted Answer

Given,
Peak value of voltage is, $V_{0} = 400 V$
Peak current is, $I_{0} = 20 A$
Frequency is, $f = 50 Hz$
For sine function, graph is increasing positively at $t = 0$
At this instant, $V = 200 \sqrt{2} V$ and $i = 10 A$

Angular frequency is, $ω = 2 π f = 2 \times π \times 50 = 100 π Hz$
Let us assume both voltage and current function is given by, $V = V_{0} \sin (ω t + ϕ_{1})$ and $i = i_{0} \sin (ω t + ϕ_{2})$
$\Rightarrow V = 400 \sin (100 π t + ϕ_{1})$
$\Rightarrow 200 \sqrt{2} = 400 \sin (ϕ_{1})$ at $t = 0$
$\Rightarrow \frac{200 \sqrt{2}}{400} = \sin (ϕ_{1})$
$\Rightarrow \sin (ϕ_{1}) = \frac{1}{\sqrt{2}}$
$\Rightarrow ϕ_{1} = 45 °$
Equation of voltage becomes, $\Rightarrow V = 400 \sin (100 π t + 45 °)$

Similarly for current,
$\Rightarrow 10 = 20 \sin (ϕ_{2})$ at $t = 0$
$\Rightarrow \sin (ϕ_{2}) = \frac{1}{2}$
$\Rightarrow ϕ_{2} = 30 °$

Equation of current becomes, $i = i_{0} \sin (100 π t + 30 °)$

Also phase difference between voltage and current is,

$ϕ = ϕ_{1} - ϕ_{2}$
$\Rightarrow ϕ = 45 ° - 30 ° = 15 °$

Now average power consumed for the AC circuit is, $P = \frac{V_{0} i_{0}}{2} \cos (ϕ)$
On substituting all the values, we have,
$\Rightarrow P = \frac{400 \times 20}{2} \times \cos (15 °) = 3864 W$

Important Questions on Alternating Current

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