Question

131) Power form a 64 V d.c. supply goes to charge a battery of 8 lead accumulators each of emf of 2.0 V sand internal resistance 1/8 $\Omega$. The charging current also runs and electric motor placed in series with the battery. If the resistance of the windings of the motor is 7.0 $\Omega$ and the steady supply current is 3.5 A, obtain
(a) the mechanical energy yielded by the miror
(b) the chemical energy, stored in the battery during charging in 1 h.

Answer 1

Dear student,

$$\begin{gathered} Terminalvoltageacrossmotor=7\times 3.5=24.5V \\ SomechanicalenergywillbeVI=24.5\times 3.5=85.75J \end{gathered}$$

$Chemicalenergy=I^{2}Rt$




$$\begin{gathered} I=\frac{E}{R+nr} \\ I=\frac{64}{7+8\times {\displaystyle \frac{1}{8}}} \\ I=8A \\ energy=I^{2}Rt=64\times 8\times \frac{1}{8}\times 1 \\ ChemicalenergyE=64J \end{gathered}$$

Regards