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  • An inductor  <img alt="\left(\mathrm{X}_{\mathrm{L}}=2 \Omega\right)" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%28%5Cmathrm%7BX%7D_%7B%5Cmathrm%7BL%7D%7D%3D2%20%5COmega%5Cri

An inductor  \left(\mathrm{X}_{\mathrm{L}}=2 \Omega\right) , a capacitor \left(X_c=8 \Omega\right) and a resistance (\mathrm{R}=8 \Omega) are connected in series with an AC source. The voltage output of AC source is given by \mathrm{V}=10 \cos (100 \pi \mathrm{t})
                  .
The instantaneous potential difference between points A and B, when the applied voltage is 3/5th of the maximum value of applied voltage is:

Option: 1

0 V


Option: 2

6 V


Option: 3

8 V


Option: 4

None of these


Answers (1)

best_answer

\mathrm{\begin{aligned} & \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^2}=10 \Omega \\ & \cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{4}{5} \\ & \therefore \quad \phi=37^{\circ} \end{aligned} }

\mathrm{\begin{array}{r} \mathrm{I}_0=\frac{\mathrm{V}_0}{\mathrm{Z}}=\frac{10}{10}=1 \mathrm{~A} \\ \therefore \quad \mathrm{I}=1 \cos \left(100 \pi \mathrm{t}+37^{\circ}\right) \\ \mathrm{V}=10 \cos (100 \pi \mathrm{t}) \end{array} }
The applied voltage is \mathrm{ \frac{3}{5} th } of the maximum applied voltage
\mathrm{ \begin{aligned} & \text { when, or } \quad 100 \pi \mathrm{t}=53^{\circ} \\ & \therefore \quad \mathrm{I}=1 \cos \left(53^{\circ}+37^{\circ}\right)=0 \\ & \therefore \quad \mathrm{V}_{\mathrm{R}}=0 \\ & \therefore \quad \mathrm{V}_{\mathrm{AB}}=\mathrm{V}_{\text {appliced }}=\frac{3}{5} \times 10=6 \mathrm{~V} \end{aligned} }

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