# The internal resistance of a battery with an EMF of 3.6 V is 0.1 ohm. Three light bulbs with a resistance of 1.5

**The internal resistance of a battery with an EMF of 3.6 V is 0.1 ohm. Three light bulbs with a resistance of 1.5 ohms each are connected in parallel to the battery. Find the potential difference across the battery terminals.**

Let’s find the load resistance of the battery Rн, if these are three lamps with a resistance Rl = 1.5 Ohm, connected in parallel:

Rn = Rl / 3 = 1.5 / 3 = 0.5 Ohm.

Let’s find the total resistance Ro of the battery and the load, if the internal resistance of the battery r = 0.1 Ohm:

Ro = Rn + r = 0.5 + 0.1 = 0.6 Ohm.

Let’s find the current flowing in the battery-load circuit, if the EMF of the battery Ub = 3.6 V:

I = Ub / Ro = 3.6 / 0.6 = 6 A.

Let’s find the voltage drop ∆U across the internal resistance of the battery:

∆U = I * r = 6 * 0.1 = 0.6 V.

Let’s find the potential difference at the ends of the battery U:

U = Ub – ∆U = 3.6 – 0.6 = 3 V.

Answer: The potential difference at the ends of the battery is 3 V.