Question

A child’s electronic toy is supplied by three 1.58-V alkaline cells having internal resistances of 0.0200Ω in series with a 1.53-V carbon-zinc dry cell having a 0.100-Ω internal resistance. The load resistance is 10.0Ω. How much power is supplied to the load?

a) 1.58 W
b) 2.0 W
c) 3.0 W
d) 4.0 W

Answer 1

The power supplied to the load in the given circuit is 3.74 W.

Step-by-step explanation:

First, let's calculate the total voltage of the battery pack by summing up the voltages of the alkaline cells and the carbon-zinc dry cell:

Total voltage = 3 * 1.58 V + 1.53 V = 6.27 V

To find the current flowing through the load, we need to calculate the total internal resistance of the battery pack. Since the alkaline cells are in series, their total internal resistance is given by:

Total internal resistance of alkaline cells = 3 * 0.0200 Ω = 0.0600 Ω

The total internal resistance of the battery pack is the sum of the internal resistance of the alkaline cells and the internal resistance of the carbon-zinc dry cell:

Total internal resistance = 0.0600 Ω + 0.100 Ω = 0.160 Ω

Now, we can calculate the current flowing through the load using Ohm's Law:

Current = Voltage / Total resistance = 6.27 V / (10.0 Ω + 0.160 Ω) = 0.610 A

Finally, we can calculate the power supplied to the load using the formula:

Power = Current^2 * Load resistance = (0.610 A)^2 * 10.0 Ω = 3.74 W

Therefore, the power supplied to the load is 3.74 W.