What is the voltage drop across R1?

Question

What is the voltage drop across R1 if a series circuit has

1. a battery with an emf of 1.5 V and an internal resistance of 3.0 Ω
2. R2 is 8.0 Ω
3. R3 is 6.0 Ω

Also, how do I calculate the terminal voltage of the battery?

Answer 1

Think of the internal resistance as just another resistor in the series. I believe that is referred to as R1 in the problem. Since the total resistance of a series circuit is the sum of the individual resistances, that would be 17.0 Ω. By Ohm's law, the current in the circuit is 1.5 v / 17.0 Ω ≈ 0.088 amperes.

Ohm's law (in form V=IR) is used again to calculate the voltage drop across each individual resistor. For R1,

V = (0.088 A) (3.00 Ω) = 0.26 v

The terminal voltage is 1.5 v - 0.26 v ≈ 1.2 v