Question

The current through the battery and resistors 1 and 2 in the figure on the left is 2.00 A. Energy is transferred from the current to thermal energy E_th in both resistors. Curves 1 and 2 in the right-hand figure give that thermal energy E_th for resistors 1 and 2, respectively, as a function of time t. The vertical scale is set by E_th,s = 80.0 mJ, and the horizontal scale is set by t_s = 30.0 s. What is the power of the battery?

Answer 1

Concepts and reason

The concepts required to solve the given problem are mechanical power, and electrical power.

Initially, calculate the power through each resistor by using expression for mechanical power and by knowing thermal energy and time for each resistor from the graph (b). Finally, take the sum of powers through each resistor to find the power of the battery.

Fundamentals

The mechanical power P is defined as the amount of energy E utilized in time t.

$P = \frac{E}{t}$

The electric power P is equal to the product of voltage V and current I.

$P = VI$

The current through each resistor is same when the resistors are connected in series and voltage is equal to the sum of voltages across each resistor.

Use the definition of mechanical power to find the power through each resistor.

The thermal power ${P_1}$through resistor ${R_1}$is given as follows:

${P_1} = \frac{{{E_{th1}}}}{{{t_s}}}$

Here, ${E_{th1}}$is the thermal energy of resistor ${R_1}.$

Substitute $80.0\,{\rm{mJ}}$for ${E_{th1}},$and 30.0 s for ${t_s}$in the above equation.

$\begin{array}{c}\\{P_1} = \frac{{80.0\,{\rm{mJ}}\left( {\frac{{1\,{\rm{J}}}}{{1000\,{\rm{mJ}}}}} \right)}}{{30.0\,{\rm{s}}}}\\\\ = 2.66 \times {10^{ - 3}}\,{\rm{W}}\\\end{array}$

The thermal power ${P_2}$through resistor ${R_2}$is given as follows:

${P_2} = \frac{{{E_{th2}}}}{{{t_s}}}$

Here, ${E_{th2}}$is the thermal energy of resistor ${R_2}.$

Substitute $40.0\,{\rm{mJ}}$for ${E_{th2}},$and 30.0 s for ${t_s}$in the above equation.

$\begin{array}{c}\\{P_2} = \frac{{40\,{\rm{mJ}}\left( {\frac{{1\,{\rm{J}}}}{{1000\,{\rm{mJ}}}}} \right)}}{{30.0\,{\rm{s}}}}\\\\ = 1.33 \times {10^{ - 3}}\,{\rm{W}}\\\end{array}$

The power P of the battery is equal to the sum of power ${P_1}$ of resistor ${R_1}$and power ${P_2}$of resistor ${R_2}.$

$P = {P_1} + {P_2}$

Substitute $2.66 \times {10^{ - 3}}\,{\rm{W}}$for ${P_1},$and $1.33 \times {10^{ - 3}}\,{\rm{W}}$for ${P_2}$in the above equation.

$\begin{array}{c}\\P = \left( {2.66 \times {{10}^{ - 3}}\,{\rm{W}}} \right) + \left( {1.33 \times {{10}^{ - 3}}\,{\rm{W}}} \right)\\\\ = 3.99 \times {10^{ - 3}}\,{\rm{W}}\left( {\frac{{1000\;{\rm{mW}}}}{{1\,{\rm{W}}}}} \right)\\\\ = 3.99\,{\rm{mW}}\\\end{array}$

Round off the value to one significant figure, the power of the battery is 4mW.

Ans:

The power of the battery is ${\bf{4}}\,{\bf{mW}}.$