Question

An alternating voltage of $260volt$ and $\omega =100rad/s$, is applied in an LCR series circuit where $L=0.01H$, $C=4\times {10}^{-4}F$, and $R=10\Omega$. The power supplied by the source is:

• A. $6760W$
• B. $3380W$
• C. $1000W$
• D. $3000W$

Answer 1

The correct option is C

$1000W$

Step 1: Given

Alternating voltage: $V_S=260V$

Angular frequency: $\omega =100rad/sec$

Inductance: $L=0.01H$

Capacitance: $C=4\times {10}^{-4}F$

Resistance: $R=10\Omega$

Step 2: Formula Used

$Z=\sqrt{{\left(X_C-X_L\right)}^2+R^2}$

$i_{rms}=\frac{V_S}{Z}$

$P={i_{rms}}^{2}R$

Step 3: Find the impedance

Calculate the inductive reactance by multiplying the angular frequency by inductance.

$$\begin{gathered} X_L=\omega L \\ =100\times 0.01 \\ =1\Omega \end{gathered}$$

Calculate the inductive capacitance by reciprocating the product of capacitance and angular frequency.

$$\begin{gathered} X_C=\frac{1}{\omega C} \\ =\frac{1}{100\times 4\times {10}^{-4}} \\ =\frac{100}{4} \\ =25\Omega \end{gathered}$$

Calculate impedance using the formula.

$$\begin{gathered} Z=\sqrt{{\left(25-1\right)}^2+{10}^2} \\ =\sqrt{{24}^2+{10}^2} \\ =\sqrt{576+100} \\ =\sqrt{676} \\ =26\Omega \end{gathered}$$

Step 4: Find the power supplied

Calculate the root mean square value of the current by dividing the voltage by impedance.

$$\begin{gathered} i_{rms}=\frac{V_S}{Z} \\ =\frac{260V}{26\Omega } \\ =10A \end{gathered}$$

Calculate the using the formula.

$$\begin{gathered} P={i_{rms}}^{2}R \\ ={\left(10A\right)}^{2}10\Omega \\ =1000W \end{gathered}$$

Hence, Option (C) is correct. the power supplied is $1000W$.