The efficiency of a cell with internal resistance of 2Ω Ω supply current to a 6Ω Ω resistor is

The efficiency of a cell with internal resistance of 2Ω $\mathrm{\Omega }$ supply current to a 6Ω $\mathrm{\Omega }$ resistor is 

Answer Details

To determine the efficiency of a cell with an internal resistance of 2 Ω while supplying current to a 6 Ω resistor, we can use the concept of power dissipation. Efficiency in this context is the ratio of the power delivered to the external resistor to the total power supplied by the cell. It can be calculated using the formula:

Efficiency (%) = (Power across load resistor / Total power output by cell) × 100

Let's break it down step by step:

• Step 1: Total Resistance - The total resistance in the circuit is the sum of the internal resistance and the resistance of the external resistor:
• Total Resistance (R_total) = Internal Resistance (R_internal) + External Resistance (R_load)
• R_total = 2 Ω + 6 Ω = 8 Ω
• Step 2: Current Calculation - The current flowing through the circuit can be calculated using Ohm's Law, which states V = IR. Assume the electromotive force (EMF) of the cell is E volts:
• Current (I) = E / R_total
• I = E / 8 (Note: We keep E as a variable as it cancels out in the efficiency equation)
• Step 3: Power Delivered to Load - The power delivered to the external resistor can be calculated as:
• Power_load = I² × R_load
• Power_load = (E/8)² × 6
• Step 4: Total Power Supplied by the Cell - The total power supplied by the cell is given by:
• Power_total = EMF × Current = E × (E/8)
• Power_total = E² / 8
• Step 5: Calculate Efficiency: Finally, plug in the power values into the efficiency formula:
• Efficiency (%) = [(E/8)² × 6 / (E² / 8)] × 100
• Simplify to Efficiency (%) = [(6/8)] × 100 = 0.75 × 100 = 75%

The efficiency of the cell when supplying current to a 6 Ω resistor with an internal resistance of 2 Ω is 75%.