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Current in an ac circuit is given by i = 3 sin $\omega$ t + 4 cos $\omega$ t then :
Option: 1
rms value of current is 5 A.

Option: 2
mean value of this current in one half period will be $\frac{6}{\pi}$

Option: 3
if voltage applied is V = V_m sin $\omega$ t then the circuit must be containing resistance and capacitance.

Option: 4
if voltage applied is V = V_m sin $\omega$ t, the circuit may contain resistance and inductance.

Answers (1)

best_answer

R C Circuit Voltage -

$V_{R}=IR$

$V_{C}={I}X_{C}$

- wherein

R.L Circuit Voltage -

$V_{R}= IR$

$V_{L}= {I}X_{L}$

- wherein

Root mean square i_{rms} -

$i_{rms}= \sqrt{\frac{i_{1}^{2}+i_{1}^{2}+----}{\eta }}= \sqrt{\overline{i}^{2}}$

$\sqrt{\frac{\int_{0}^{T}i^{2}dt}{\int_{0}^{T}dt}}= \frac{{i_{0}}'}{\sqrt{2}}= 0.707\, {i}'_{0}$

$= 70.7$ % ${i}'_{0}$

- wherein

${i}'_{0}$ - Peak Current

i = 3 sin $\omega$ t + 4 cos $\omega$ t

$= 5\left [ \frac{3}{5}sin \omega t+\frac{4}{5}\cos \omega t \right ]$

$= 5\left [ \sin (\omega t +\delta ) \right ]$ ....................(1)

rms value $= \frac{5}{\sqrt{2}}\Rightarrow$ mean value $=\frac{\int_{T_{1}}^{T_{2}}idt}{\int_{T_{1}}^{T_{2}}dt}$

$\therefore$ Initial value of time is not given hence the mean value will be different for various time intervals.

If voltage applied is V = V_m sin $\omega$ t then i given by equation (1) indicates that it is ahead of V by $\delta$ where 0 < $\delta$ < 90 which indicates that the circuit contains R & C.

Posted by

Ajit Kumar Dubey

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