Question

The dry cell of E.M.F $12V$, is connected in series to a $3\Omega$ and a $6\Omega$ resistor. Calculate the heat energy produced by the $3\Omega$ resistor in $5s$.

• A. $\frac{80}{3}J$
• B. $\frac{135}{16}J$
• C. $80J$
• D. $480J$

Answer 1

The correct option is A $\frac{80}{3}J$

EMF of the cell $V=12V$

Total Resistance, $R_T=3+6=9\Omega$

Therefore, Current passing through the resistors is $i=\frac{V}{R_t}=\frac{12}{9}=\frac{4}{3}A$

Power across $3\Omega ,$$P=i^{2}R=\frac{16}{9}\times 3=\frac{16}{3}W$

Heat energy produced, $E=P\times t=\frac{16}{3}\times 5=\frac{80}{3}J$