a chemical cell of internal resistance 1\(\Omega\) supplies electric current to an external resistor of resistance 3\(\Omega\). Calculate the efficiency of ...

The efficiency of a cell is defined as the ratio of the electrical energy output to the chemical energy input. In other words, it is the ratio of the useful output power to the total input power. The useful output power is the power delivered to the external load, while the total input power is the sum of the useful output power and the power dissipated in the internal resistance of the cell. In this case, the external resistor has a resistance of 3\(\Omega\), and the internal resistance of the cell is 1\(\Omega\). The current flowing through the circuit can be calculated using Ohm's law: I = V / R where I is the current, V is the voltage, and R is the total resistance of the circuit. The total resistance of the circuit is the sum of the external resistance and the internal resistance: R_total = R_external + R_internal = 3\(\Omega\) + 1\(\Omega\) = 4\(\Omega\) The voltage across the external resistor can be calculated using Ohm's law again: V_external = I * R_external The current flowing through the circuit can be calculated using Ohm's law: I = V / R_total where V is the voltage across the cell and the external resistor. The power delivered to the external resistor is given by: P_external = V_external * I The power dissipated in the internal resistance is given by: P_internal = I^2 * R_internal The total input power is the sum of the useful output power and the power dissipated in the internal resistance: P_input = P_external + P_internal The efficiency of the cell is given by: efficiency = P_external / P_input Substituting the expressions for P_external, P_internal, and P_input, we get: efficiency = V_external * I / (V_external * I + I^2 * R_internal) Simplifying this expression and substituting the values given in the problem, we get: efficiency = 3 / 4 = 0.75 or 75% Therefore, the correct answer is 75%.