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

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

• Imagine the circuit as having an ideal ammeter in series with a resistor. You still know the current through the bulb and the voltage drop across it. You could still calculate the resistance of the bulb as you have outlined. Commented Apr 11 at 19:51
• What resistance “across the bulb”? The ammeter resistance is in series with the bulb. Did you mean the voltage across the bulb? Commented Apr 11 at 20:57

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