Question

Question: (a) In Fig. 27-56, what current does the ammeter read if\(\varepsilon = 5.0\;{\rm{V}}\)(ideal battery),\({R_1}\; = \;2.0\;{\rm{\Omega }}\), \({R_2}\; = \;4.0\;{\rm{\Omega }}\), and \({R_3}\; = \;6.0\;{\rm{\Omega }}\)? (b) The ammeter and battery are now interchanged. Show that the ammeter reading is unchanged.

Short answer

Question: Integrated Concepts

Hydroelectric generators (see Figure 20.43) at Hoover Dam produce a maximum current of \({\rm{8}}{\rm{.00\times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;A}}\)at \({\rm{250 kV}}\). (a) What is the power output? (b) The water that powers the generators enters and leaves the system at low speed (thus its kinetic energy does not change) but loses \({\rm{160\;m}}\) in altitude. How many cubic meters per second are needed, assuming \({\rm{85}}{\rm{.0\% }}\) efficiency?

(a) The output power is \({\rm{2}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{\;W}}\).

(b) The required solution is \(1.50\times{10^3}{\rm{{{m^3}} \mathord{\left/

{\vphantom {{{m^3}} s}} \right.

\kern-\nulldelimiterspace} s}\).

Step-by-step solution

Definition of Power:

The power of a process is the amount of some type of energy converted into a different type divided by the time interval\({\rm{\Delta t}}\)in which the process occurred:

\({\rm{P}} &= \frac{{{\rm{\Delta E}}}}{{{\rm{\Delta t}}}}\) ….. (1)

The SI unit of power is the watt.

As you know, one watt is equal to one joule per second. Therefore,

\(1{\rm{ }}W &= 1{\rm{ }}{J \mathord{\left/

{\vphantom {J s}} \right.

\kern-\nulldelimiterspace} s}\)

The power of any electric circuit is defined by the formula

\({\rm{P}} &= {\rm{I\Delta V}}\) ….. (2)

Here,\({\rm{I}}\)is the current and\({\rm{I\Delta V}}\)is the potential difference.

123

The gravitational potential energy is given by

\({{\rm{U}}_{\rm{g}}} &= {\rm{mgh}}\) ….. (3)

Here,\({\rm{m}}\)is the mass of the particle,\({\rm{g}}\)is the acceleration due to gravity and\({\rm{h}}\)is the height of the particle.

xy

The maximum current produced by the generators, \({\rm{I}} &= {\rm{8}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;A}}\).

The voltage produced by the generators is,

\(\begin{aligned}{c}{\rm{\Delta V}} &= \left( {{\rm{250 kV}}} \right)\left( {\frac{{{\rm{1000\;V}}}}{{{\rm{1 kV}}}}} \right)\\ &= {\rm{250 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;V}}\end{aligned}\).

The kinetic energy of the water is constant.

The change in height of the water,\({\rm{\Delta y}} &= {\rm{160\;m}}\).

The efficiency of the energy conversion process, \({\rm{\eta }} &= {\rm{85}}{\rm{.0\% }}\).

The density of water, \({\rm{\rho }} &= {\rm{1000 }}{{kg} \mathord{\left/

{\vphantom {{kg} {{m^3}}}} \right.

\kern-\nulldelimiterspace} {{m^3}}}\).

xyz

The output power of the generator is,

\({\rm{P}} &= {\rm{I\Delta V}}\)

Substituting the known values in the above equation.

\(\begin{aligned}{c}{\rm{P}} &= \left( {{\rm{8}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;A}}} \right)\left( {{\rm{250 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;V}}} \right)\\ &= {\rm{2}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{\;W}}\end{aligned}\)

Therefore, the output power is given by \({\rm{2}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{\;W}}\).