10. A cell of internal resistance 2 ohms supplies
current to a 6-ohm resistor. The efficiency of the
cell is
A. 12.0 %
B. 25.0 %
C. 33.3 %
D. 75.0 %
To find the efficiency of the cell, we can use the formula:
Efficiency = (Power output / Power input) * 100%
Power output is the power dissipated in the 6-ohm resistor, which can be calculated using Ohm's Law:
Power output = I^2 * R
Power input is the power supplied by the cell, which can be calculated using the formula:
Power input = I^2 * (r + R)
where r is the internal resistance of the cell.
Given that the internal resistance is 2 ohms and the external resistance is 6 ohms, we can calculate the power output and power input:
Power output = I^2 * R = (I^2)(6)
Power input = I^2 * (r + R) = (I^2)(2 + 6)
To find the efficiency, we divide the power output by the power input and multiply by 100%:
Efficiency = [(I^2)(6) / (I^2)(2 + 6)] * 100%
Simplifying the equation:
Efficiency = (6 / 8) * 100%
Efficiency = 0.75 * 100%
Efficiency = 75%
Therefore, the efficiency of the cell is 75% or option D.
To find the efficiency of the cell, we can use the formula:
Efficiency = (Useful output energy / Total input energy) x 100
In this case, the useful output energy is the power dissipated by the resistor, and the total input energy is the power supplied by the cell.
The power dissipated by the resistor can be calculated using the formula:
Power = (Current)^2 x Resistance
Given that the internal resistance of the cell is 2 ohms and the resistance of the resistor is 6 ohms, the total resistance in the circuit is 2 + 6 = 8 ohms.
To find the current flowing in the circuit, we use Ohm's Law:
V = I x R
The voltage supplied by the cell is equal to the voltage across the internal resistance plus the voltage across the resistor:
V = I x (2 + 6)
The voltage across the internal resistance can be calculated using Ohm's Law:
V_internal = I x 2
Substituting the values, we have:
V = I x (2 + 6) = 8I
V_internal = I x 2
The voltage supplied by the cell is equal to the sum of the voltage across the internal resistance and the voltage across the resistor:
8I = 2I + 6I
8I = 8I
Hence, the current flowing in the circuit is I = 1A.
Now, we can calculate the power dissipated by the resistor:
Power = (Current)^2 x Resistance = (1)^2 x 6 = 6W
The power supplied by the cell can be calculated using the formula:
Power = Voltage x Current
Since we know that the voltage supplied by the cell is 8I = 8V and the current flowing in the circuit is I = 1A, we have:
Power = 8V x 1A = 8W
Now, substituting the values into the efficiency formula:
Efficiency = (Useful output energy / Total input energy) x 100
= (6W / 8W) x 100
= 0.75 x 100
= 75.0%
Therefore, the efficiency of the cell is D. 75.0%.