How would having a non-ideal ammeter affect the resistance across this bulb?

Question

A non-ideal ammeter is one that has resistance right. So that means that work will be done by the current flowing through the ammeter and there will be a potential difference across it. This will then decrease the potential difference across the bulb as a result (my understanding of this come from Kirchhoff's 2nd Law; which state that around any closed loop in a circuit the sum of the cells potentials equals the sum on the component potentials). The current has decreased also as there is more resistance in the circuit and that the voltage has decreased. So would the calculated resistance remain the same and not change? and so do the voltage and current decrease by the same factor in order for this to happen? Im not sure could someone clarify

Answer 1

Adding the ammeter resistance means less current is drawn from the battery and the measured voltage across the bulb will decrease but not the resistance of the bulb.

So the ammeter without resistance will provide measure a higher current than the ammeter with resistance.

Let the battery provide 120V, the resistance of the light bulb $R_{bulb} = 10\Omega$, and the resistance of the ammeter $R_{meter} = 2\Omega$.

Without the ammeter resistance:

$$V = IR_{bulb}$$ Substituting values gives $I = 12 $ Amps which is measured by the ammeter and the measured voltage across the bulb $V_{bulb} = 120 $ volts

With the ammeter resistance: $$V = I_2(R_{meter} + R_{bulb})$$ Substituting values gives $I_2 = 10$ Amps and $V_{bulb} = 100$ Volts.