An electric current of \( 4\,\mathrm{A} \) flows through a \( 12\,\Omega \) resistor as shown in the figure. The rate of heat energy produced in the resistor \( 8\,\Omega \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{128}{9}\,\mathrm{W} \)
The whole 4 A flows through the \( 12\,\Omega \) resistor and then divides between the \( 8\,\Omega \) and \( 4\,\Omega \) resistors, which are in parallel. The current divides in the inverse ratio of the resistances.
\( \displaystyle I_8 = I\times\frac{R_4}{R_8 + R_4} = 4\,\mathrm{A}\times\frac{4\,\Omega}{12\,\Omega} = \frac{4}{3}\,\mathrm{A} \)
The rate of heat produced is the Joule heating \( P = I^{2}R \):
\( \displaystyle P_8 = \left(\frac{4}{3}\,\mathrm{A}\right)^{2}\times8\,\Omega = \frac{16}{9}\times8\,\mathrm{W} = \frac{128}{9}\,\mathrm{W} \)