Question

Six identical elements are connected in a battery whose internal resistance is 6Ω. Resistors R₁=14Ω and R₂ are connected to the battery, the total resistance of the external circuit is R=7Ω and the current flowing in it is 4 A. Determine the resistance of the second resistor and the internal EDS of the elements .

Answer 1

Using Ohm's Law, we can calculate the voltage of the battery as V = IR = (4 A)(7 Ω) = 28 V.

The total resistance of the circuit is R = R₁ + R₂ + r, where r is the internal resistance of each element.

We know that R₁ = 14 Ω and R = 7 Ω, so we can solve for R₂:

R₂ = R - R₁ = 7 Ω - 14 Ω = -7 Ω

This is a negative resistance, which doesn't make sense physically. However, it indicates that there is an error in the problem or the calculations.

To find the internal EDS of the elements, we can use the equation:

V = ε - Ir

where V is the voltage of the battery, ε is the internal EDS of each element, I is the current flowing in the circuit, and r is the internal resistance of each element.

We know that V = 28 V, I = 4 A, and r = 6 Ω, so we can solve for ε:

ε = V + Ir = 28 V + (4 A)(6 Ω) = 52 V

Therefore, the internal EDS of each element is 52 V.