A current of −8 + 6√2 (sinωt + 30°) A is passed through three meters. These are a zero centered PMMC meter, a true RMS meter, and a moving iron instrument. The respective readings (in A) will be
A current of −8 + 6√2 (sinωt + 30°) A is passed through three meters. These are a zero-centered PMMC meter, a true RMS meter, and a moving iron instrument. The respective readings (in A) will be
Right Answer is:
−8, 10, 10
SOLUTION
⇒ PMMC reads only DC value
A waveform that consists of both DC & AC components can be represented as
Ao + A(sinωt + φ)
Where
Ao = DC Component
A(sinωt + φ) = AC component (instantaneous value & A is peak value)
Since PMMC reads only DC value
PMMC Reading = −8 A
Moving iron meter and RMS meter reads RMS value of current therefore
$\begin{array}{l}{I_{RMS}} = \sqrt {A_o^2 + {{\left( {\dfrac{A}{{\sqrt 2 }}} \right)}^2}} \\\\{I_{RMS}} = \sqrt { – {8^2} + {{\left( {\dfrac{{6\sqrt 2 }}{{\sqrt 2 }}} \right)}^2}} \end{array}$
IRMS = 10 A
RMS Meter Reading = 10 A
Moving Iron Meter Reading = 10 A