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.
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