Question

Figure show parallel circuit with $R_A=8\Omega$, $X_L=6\Omega$, $R_B=6\Omega$ and $X_C=8\Omega$. The current supplied by alternating source $V$ is $I=14\text{A}.$ (take sin ${37}^0=\frac{3}{5})$

• A. Current passing through the resistor $R_A$ is $7\sqrt{2}\text{A}$
• B. Power dissipated in the circuit is $1000\text{W}$
• C. Current passing through the resistor $R_A$ is $10\sqrt{2}\text{A}$
  
• D. Power dissipated in the circuit is $1372\text{W}$

Answer 1

Answer: (A), (D)

Power dissipated in the circuit is $1372\text{W}$


From the figure, $I_A=\frac{V}{10}$ and $I_B=\frac{V}{10}$


$\therefore I=\sqrt{I_A^2+I_B^2}$

$\Rightarrow V=\frac{140}{\sqrt{2}}=70\sqrt{2}\text{V}$

$\Rightarrow I_A=I_B=7\sqrt{2}....\left(1\right)$

Total power dissipated in the circuit , $P=I_A^2{R}_A+I_B^2{R}_B$

$=(7\sqrt{2}{)}^2\times 8+(7\sqrt{2}{)}^2\times 6$
$=14\times 49\times 2$
$=1372\text{W}$