Question

Total instantaneous power supplied by 3-phase ac supply to a balanced R-L load is

• A. zero
• B. constant
• C. pulsating with zero average
• D. pulsating with non-zero average

Answer 1

The correct option is B constant

Impedance of the load
$=Z_L=R+j\omega L=|Z|\angle {\theta }_L$
where, ${\theta }_L=ta{n}^{-1}\left(\frac{\omega L}{R}\right)$
Voltages of $3-\varphi$ supply
$V_a=V_{m}sin\omega t$
$V_b=V_{m}sin(\omega t-{120}^{\circ })$
$V_c=V_{m}sin(\omega t+{120}^{\circ })$
$I_a=\frac{V_a}{Z_L}=\frac{V_{m}sin\omega t}{|Z|\angle {\theta }_L}=I_{m}sin(\omega t-{\theta }_L)$

where, $I_m=\frac{V_m}{|Z|}$

Similarly,
$I_b=I_{m}sin(\omega t-120-{\theta }_L)$ and
$I_c=I_{m}sin(\omega t+120-{\theta }_L)$

Instantaneous power, P $=V_a{I}_a+V_b{I}_b+Vc{I}_c$
$P=V_m{I}_m[sin\omega t.sin(\omega t-{\theta }_L+sin(\omega t-{120}^{\circ })$
$sin(\omega t-120-{\theta }_L)+sin(\omega t+{120}^{\circ })$
$sin(\omega t+120-{\theta }_L)]$

$=\frac{V_m{I}_m}{2}\left[\{cos{\theta }_L-cos(2\omega t-{\theta }_L\left)\right\}\right]$
$+\{cos{\theta }_L-cos(2\omega t-240-{\theta }_L\left)\right\}$
$+\{cos{\theta }_L-cos(2\omega t+240-{\theta }_L\left)\right\}$

$\Rightarrow P=\frac{3V_m{I}_m}{2}cos\varphi =constant$