Question

A $50\,{\text{V}}$ d.c. a power supply is used to charge a battery of eight lead accumulators, each of emf $2\,{\text{V}}$ and internal resistance $1/8\,\Omega$. The charging current also runs a motor connected in series with the battery. The resistance of the motor is $5\,\Omega$ and the steady current supply is $4\,{\text{A}}$. The mechanical power stored in the motor is
A. $80\,{\text{W}}$
B. $40\,{\text{W}}$
C. $64\,{\text{W}}$
D. $30\,{\text{W}}$

Answer 1

Use the equation for the power. Determine the total power supplied by the d.c. supply. Then determine the power lost as heat to the motor and the accumulators (battery) and the unusable power from the battery. Determine the total chemical power due to 8 accumulators stored in the batter and then take the difference of the unusable power and the mechanical power to determine the mechanical power stored in the motor.

Formulae used:
The equation for the power is given by
$P = {I^2}R$ …… (1)
Here, $P$ is the power, $I$ is the current and $R$ is the resistance.
Another equation for the power is given by
$P = IV$ …… (2)
Here, $P$ is the power, $I$ is the current and $V$ is the potential difference.

Complete step by step answer:
It is given that the d.c. power supply is of $50\,{\text{V}}$ and the steady current is $4\,{\text{A}}$.
We can determine the total power supplied by the d.c. battery using equation (1).
Substitute $4\,{\text{A}}$ for $I$ and $50\,{\text{V}}$ for $V$ in equation (2).
$P = \left( {4\,{\text{A}}} \right)\left( {50\,{\text{V}}} \right)$
$\Rightarrow P = 200\,{\text{W}}$
Hence, the total power supplied by the supply is $200\,{\text{W}}$.
Some of the power supplied by the d.c. supply is lost as heat to the accumulators and motor.
There are 8 accumulators and the resistance of each accumulator is $1/8\,\Omega$.
We can determine the power lost ${P_a}$ to the accumulators.
The power lost to one accumulator is ${I^2}R$ and to 8 accumulators is $8{I^2}R$.
Substitute $4\,{\text{A}}$ for $I$and $1/8\,\Omega$ for $R$ in the above equation.
${P_a} = 8{\left( {4\,{\text{A}}} \right)^2}\left( {\dfrac{1}{8}\,\Omega } \right)$
${P_a} = 16\,{\text{W}}$

The resistance of the motor is $5\,\Omega$.
$r = 5\,\Omega$
Now determine the power lost to the motor.
Rewrite equation (1) for the power lost ${P_m}$ to the motor.
${P_m} = {I^2}r$

Substitute $4\,{\text{A}}$ for $I$and $5\,\Omega$ for $r$ in the above equation.
${P_m} = {\left( {4\,{\text{A}}} \right)^2}\left( {5\,\Omega } \right)$
$\Rightarrow {P_m} = 80\,{\text{W}}$

The power ${P_u}$ which is not used is the difference of the total power supplied and the sum of the power lost as heat.
${P_u} = P - \left( {{P_a} + {P_m}} \right)$
$\Rightarrow {P_u} = 200\,{\text{W}} - \left( {16\,{\text{W}} + 80\,{\text{W}}} \right)$
$\Rightarrow {P_u} = 104\,{\text{W}}$

We can now calculate the total chemical power ${P_c}$ stored in the battery which is the product of the emf of 8 accumulators and the steady current.
${P_c} = I\left( {8{V_a}} \right)$

Substitute $4\,{\text{A}}$ for $I$and $2\,{\text{V}}$ for ${V_a}$ in the above equation.
${P_c} = \left( {4\,{\text{A}}} \right)\left( {8\left( {2\,{\text{V}}} \right)} \right)$
$\Rightarrow {P_c} = 64\,{\text{W}}$

Hence, the chemical power stored in the battery is $64\,{\text{W}}$.

The mechanical power ${P_{mec}}$ stored in the motor is the difference of the unusable power ${P_u}$ and the chemical power ${P_c}$ stored in the battery.
${P_{mec}} = {P_u} - {P_c}$

Substitute $104\,{\text{W}}$ for ${P_u}$ and $64\,{\text{W}}$ for ${P_c}$ in the above equation.
${P_{mec}} = \left( {104\,{\text{W}}} \right) - \left( {64\,{\text{W}}} \right)$
$\Rightarrow {P_{mec}} = 40\,{\text{W}}$

Therefore, the mechanical power stored in the motor is $40\,{\text{W}}$.

So, the correct answer is “Option B”.

Note:
The students while determining the power lost in the form of heat to the battery may substitute the values of the current and resistance directly in the formula. But the students should keep in mind that there are 8 accumulators in the battery. Hence, the power should be multiplied by 8 to determine the power lost to the battery.