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 a 3.65 Ω resistance. How much more power is delivered to the radio by alkaline cell, which has an internal resistance of 0.200Ω than by an nicad cell, having an internal resistance of 0.0.040Ω?

0.2 W more power than nicad cell is delivered by alkaline cell

Explanation:

FOR NICKEL-CADMIUM CELL (nicads):

First we find the current supplied to radio by the cell. For this purpose, we use the formula:

I = E/(R+r)

where,

I = current supplied

E = emf of cell = 1.25 V

R = resistance of radio = 3.65 Ω

r = internal resistance of cell = 0.04 Ω

Therefore,

I = (1.25 V)/(3.65 Ω + 0.04 Ω)

I = 0.34 A

Now, we calculate the power delivered to radio by following formula:

P = VI

but, from Ohm's Law:   V = IR

Therefore,

P = I²R

where,

P = Power delivered = ?

I = current = 0.34 A

R = Resistance of radio = 3.65 Ω

Therefore,

P = (0.34 A)²(3.65 Ω)

P = 0.41 W

FOR ALKALINE CELL:

First we find the current supplied to radio by the cell. For this purpose, we use the formula:

I = E/(R+r)

where,

I = current supplied

E = emf of cell = 1.58 V

R = resistance of radio = 3.65 Ω

r = internal resistance of cell = 0.2 Ω

Therefore,

I = (1.58 V)/(3.65 Ω + 0.2 Ω)

I = 0.41 A

Now, we calculate the power delivered to radio by following formula:

P = VI

but, from Ohm's Law:   V = IR

Therefore,

P = I²R

where,

P = Power delivered = ?

I = current = 0.41 A

R = Resistance of radio = 3.65 Ω

Therefore,

P = (0.41 A)²(3.65 Ω)

P = 0.61 W

Now, fo the difference between delivered powers by both cells:

ΔP = (P)alkaline - (P)nicad

ΔP = 0.61 W - 0.41 W

ΔP = 0.2 W

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete