Question

Two identical cells connected in series send 1.0 A current through a 5 Ω resistor. When they are connected in parallel, they send 0.8 A current through the same resistor. What is the internal resistance of the cell?

A. 0.5 Ω
B. 1.0 Ω
C. 1.5 Ω
D. 2.5 Ω

Answer 1

The internal resistance of the cell is 25 Ω.

Step-by-step explanation:

To find the internal resistance of the cell, we can use the equations relating current, voltage, and resistance. When the cells are connected in series, they create a voltage drop across the resistor of 5 Ω, which results in a current of 1.0 A. In this case, the total voltage supplied by the cells is the sum of the voltage drops across the resistor and the internal resistance of the cells. Let's call the internal resistance R.

Series Connection:

V = IR + 5 Ω * 1.0 A = 5 Ω + IR

When the cells are connected in parallel, they still create a voltage drop across the resistor of 5 Ω, but the total current flowing through the resistor is now 0.8 A.

Parallel Connection:

V = IR + 5 Ω * 0.8 A = 5 Ω + 0.8R

We can now solve these two equations simultaneously to find the value of R.

5 Ω + IR = 5 Ω + 0.8R

IR = 0.8R

0.2R = 5 Ω

R = 5 Ω / 0.2 = 25 Ω

Therefore, the internal resistance of the cell is 25 Ω.