A cell of internal resistance 1Ω supplies current to an external resistor of 3Ω. The efficiency of the cell is

Answer Details

The efficiency of a cell is defined as the ratio of the power output to the power input. In this case, the power output is the power dissipated in the external resistor, and the power input is the power supplied by the cell.
First, we can calculate the current in the circuit using Ohm's law:
I = V / R
where V is the voltage supplied by the cell, and R is the total resistance of the circuit. In this case, R is the sum of the internal resistance of the cell (1Ω) and the external resistor (3Ω):
R = 1Ω + 3Ω = 4Ω
Now, we can calculate the current:
I = V / R = V / 4Ω
Next, we can calculate the power output:
Pout = I^2 * Rext
where Rext is the resistance of the external resistor. In this case, Rext is 3Ω, so:
Pout = I^2 * 3Ω
Finally, we can calculate the power input:
Pin = I^2 * R
where R is the total resistance of the circuit, including the internal resistance of the cell. In this case, R is 4Ω, so:
Pin = I^2 * 4Ω
Now, we can calculate the efficiency:
efficiency = Pout / Pin
Substituting the values we calculated earlier:
efficiency = (I^2 * 3Ω) / (I^2 * 4Ω) = 3/4 = 0.75 = 75%
Therefore, the efficiency of the cell is 75%.