A cell of internal resistance of 2Ω Ω supplies current through a resistor, X if the efficiency of the cell is 75%, find the value of X.

A cell of internal resistance of 2Ω $\mathrm{\Omega }$ supplies current through a resistor, X if the efficiency of the cell is 75%, find the value of X.

Answer Details

To solve the problem, let's first understand the concept of efficiency in this context. Efficiency refers to the ratio of the useful power output to the total power output of a system. In simpler terms, it tells us how much of the power provided by the cell is being effectively used by the resistor, X.

Given that the cell has an internal resistance (r) of 2Ω and we need the efficiency to be 75%, we will follow these steps:

• The total resistance in the circuit is the sum of the internal resistance and the resistance of the resistor X, so it is (r + X).
• Efficiency is given by the formula:

Efficiency (%) = (R / (R + r)) * 100

Where:

• R is the resistance of resistor X.
• r is the internal resistance of the cell.

According to the problem, efficiency is 75%, so:

(X / (X + 2)) * 100 = 75

First, let’s eliminate the percentage by dividing both sides by 100:

(X / (X + 2)) = 0.75

Now, let's solve for X:

1. Multiply both sides by (X + 2):

X = 0.75 * (X + 2)



2. Distribute the 0.75:

X = 0.75X + 1.5



3. Subtract 0.75X from both sides:

0.25X = 1.5



4. Finally, divide by 0.25:

X = 1.5 / 0.25



5. Solve for X:

X = 6 Ω

Hence, for the cell to have an efficiency of 75%, the value of the resistor X must be 6Ω.