Question

an ac voltage of the form v=8sin2πt is applied to a circuit consists of three capacitors of capacitance 5μf,10μf and 15μf in series? find the following by considering all quantities in si units? a) AC current by) peak value of AC current c) capacitive reactance​

Answer 1

Answer:

General form of Voltage is given as:

V = V₀.Sin (wt)   ...(i)

According to the question the voltage is given as:

V = 8.Sin (2πt)  ...(ii)

Hence the value of omega (w) is found to be 2π by comparing (i) and (ii).

Now we are required to calculate:

1. AC Current
2. Peak Value or r.m.s value of Current
3. Capacitive Reactance

Now we are given 3 capacitors connected in series with each other. Hence the net capacitance will be:

$\dfrac{1}{C_{eff}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3}\\\\\\\implies \dfrac{1}{C_{eff}} = \dfrac{1}{5} + \dfrac{1}{10} + \dfrac{1}{15}\\\\\\\implies \dfrac{1}{C_{eff}} = \dfrac{6+3+2}{30}\\\\\\\implies \boxed{ \bf{C_{eff} = \dfrac{30}{11} = 2.72\:\: \mu F}}$

We know that,

$\text{Capacitive Reactance} = \dfrac{V_0}{I_0} = \dfrac{1}{\omega C}\\\\\\\implies \text{Capacitive Reactance} = \dfrac{8}{I_0} = \dfrac{1}{2\pi \times 2.72 \times 10^{-6}}\\\\\\\implies \boxed{ \bf{\textbf{Capacitive Reactance} = \dfrac{8}{I_0} = 5.85 \times 10^4\:\: \Omega}}$

Now using Capacitive Reactance, the alternating current I₀ can be found. Hence we get:

$\implies \dfrac{8}{I_0} = 5.85 \times 10^4\\\\\\\implies \dfrac{8}{5.85 \times 10^4} = I_0\\\\\\\implies I_0 = 1.36 \times 10^{-4}\\\\\\\implies \boxed{ \bf{ I_0 = 136 \:\: \mu C}}$

The formula to calculate the R.M.S. value of current is:

$\implies I_{rms} = \dfrac{I_0}{\sqrt{2}}$

Hence the peak value or the RMS value of current is:

$\implies I_{rms} = \dfrac{136}{1.414} \times 10^{-6}\\\\\\\implies \boxed{ \bf{I_{rms} = 96.18 \:\: \mu C}}$

Answer 2

Answer:

vmsinωt=qC. To find the current, we use the relation: i=dqdt. i=ddt(vmCsinωt)=ωCvmcos(ωt). Using the relation: cos(ωt)=sin(ωt+π2).

Explanation:

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