# A galvanic cell with an EMF of 1.5V and an internal resistance of 1 0m is closed to an external resistance of 4 Ohm.

**A galvanic cell with an EMF of 1.5V and an internal resistance of 1 0m is closed to an external resistance of 4 Ohm. Find the current in the circuit, the voltage drop in the internal part of the circuit, and the voltage at the terminals of the element.**

To determine the current in the circuit, we use Ohm’s law for a complete electrical circuit:

I = E / (R + r),

where E is the electromotive force, R is the external resistance of the circuit, r is the internal resistance of the EMF source, I is the current;.

Let’s calculate the current:

I = E / (R + r) = 1.5 / (4 + 1) = 1.5 / 5 = 0.3 (A).

The voltage drop in the inner part of the circuit is determined from the formula:

Ur = I * r.

Substituting the numerical values of the current strength and internal resistance, we get:

Ur = I * r = 0.3 * 1 = 0.3 (B).

The voltage drop across the terminals is determined from the formula:

UR = I * R.

Substituting the numerical values of the current strength and external resistance, we get:

UR = I * R = 0.3 * 4 = 1, 2 (B).

Answer: I = 0.5 (A), Ur = 0.3 (B), UR = 1.2 (B).