Answer:
0.2 W more power than nicad cell is delivered by alkaline cell
Explanation:
<u>FOR NICKEL-CADMIUM CELL (nicads):</u>
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
<u>FOR ALKALINE CELL:</u>
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
<u>ΔP = 0.2 W</u>
