If the rms value of the voltage of the waveform shown below is p q V , then what is the value of p + q ? ( p and q are the smallest positive integers.)

V rms 2 = 1 T ∫ 0 T V 2 dt = 2 T ∫ 0 T 2 V 2 dt T = 2 sec, V = 2t 0 ≤ t ≤ 1 V rms 2 = 2 T ∫ 0 1 4 t 2 dt = ∫ 0 1 4 t 2 dt = 4 t 3 3 ] 0 1 = 4 3 V rms = 4 3 V

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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACAC current and voltageEasy2 minPYQ_2020

PhysicsEasynumerical

If the rms value of the voltage of the waveform shown below is $\sqrt{\frac{p}{q}} V$ , then what is the value of $(p + q)$ ? ( $p$ and $q$ are the smallest positive integers.)

Answer:

7.00

Solution:

$V_{rms}^{2} = \frac{1}{T} \int_{0}^{T} V^{2} dt = \frac{2}{T} \int_{0}^{\frac{T}{2}} V^{2} dt$

T = 2 sec,

V = 2t $0 \leq t \leq 1$

$V_{rms}^{2} = \frac{2}{T} \int_{0}^{1} 4 t^{2} dt$

$= \int_{0}^{1} 4 t^{2} dt$

$= 4 \frac{t^{3}}{3}]_{0}^{1} = \frac{4}{3}$

$V_{rms} = \sqrt{\frac{4}{3}} V$

Subject:PhysicsTopic:EMI/ACSubtopic:AC current and voltage

⏱ 2mℹ️ Source: PYQ_2020

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