Answer to Question #164683 in Electric Circuits for elizabeth

Question

The label on a battery-powered radio recommends the use of a rechargeable nickel-cadmium cell (nicads),although it has a 1.25-V emf, whereas an alkaline cell has a 1.58-V emf. The radio has a3.20 Ωr esistance.(a) Draw a circuit diagram of the radio and its battery.Now, calculate the power delivered to the radio (b) when using a nicad cells, each having an internal resistance of 0.0400 Ω, and (c) when using an alkaline cell, having an internal resistance of0.200 Ω. (d) Does this difference seem significant, considering that the radio’s effective resistance is lowered when its volume is turned up

Answer 1

a)

b) Given quantites: $\epsilon = 1.25V$ $r = 0.04 \Omega$ $r = 3.2\Omega$

Ohm's law: $U = IR$

The terminal voltage of the battery: $U = \epsilon -Ir$

So $IR = \epsilon -Ir$

$I = \large\frac{\epsilon}{R+r}$ $= \large\frac{1.25}{3.2+0.04}$ $=0.386A$

Electric power: $P = UI = I^2R = (0.386)^2*3.2 = 0.477W$

c) Given quantites: $\epsilon= 1.58V$ $r = 0.2\Omega$ $R = 3.2\Omega$

Ohm's law: $I = \large\frac{\epsilon}{R+r}$ $= \frac{1.58}{3.2+0.2} = 0.465A$

Electric power: $P = (0.465)^2*3.2 = 0.692W$

d) There is no significant power difference between nickel-cadmium cell and alkaline cell (0.2W).

Specially that when the radio volume is turned up, its effective resistence is lowered then this power diiference will decreases more than that