A cell whose internal resistance is… | Physics

Question 1 Report

A cell whose internal resistance is \(0.55\,\Omega\) delivers a current of 4 A to an external resistor. The lost voltage of the cell is?

4.00 V

2.20 V

0.15 V

8.00 V

Answer Details

The lost voltage of the cell can be calculated using Ohm's Law, which states that voltage (V) is equal to the current (I) multiplied by the resistance (R) of the circuit, plus the voltage lost within the internal resistance (r) of the cell. So, we can use the formula V = IR + Ir to find the lost voltage of the cell, where I is the current (4 A), R is the external resistance (unknown), and r is the internal resistance of the cell (0.55 Ω). Rearranging the formula, we get V - IR = Ir. Substituting the values we know, we get: V - (4 A x R) = 4 A x 0.55 Ω Simplifying this equation, we get: V - 4R = 2.2 V So, the lost voltage of the cell is 2.2 volts. Therefore, the correct option is 2.20 V.

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Current Electricity

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A cell whose internal resistance is \(0.55\,\Omega\) delivers a current of 4 A to an external resistor. The lost voltage of the cell is?

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