An ammeter has a resistance of 1.0. The circuit consists of a 10. Resistor that is in series with a 6.0 V ideal battery...?

(a)
Refer to the circuit diagram below.

Resistance of the ammeter
= 1.0 mΩ
= 1.0/1000 Ω
= 0.001 Ω

Equivalent resistance of the circuit, Rₑ
= (10. + 0.001) Ω
= 10.001 Ω

Current in the ammeter, I
= V/Rₑ
= 6/10.001
= 0.59994 A

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(b)
Voltage drop across the 10. Ω resistor
= I R
= 0.59994 × 10
= 5.9994 V

OR:
Voltage drop across the 10. Ω resistor
= 6 × (R/Rₑ)
= 6 × 10/10.001
= 5.9994 V

THe current is found using Ohms law   V = IR

Here R = 10 Ohm + 0.001 Ohm = 10.001 Ohms

V = 6 V so  I = 0.59994 Amps

THe voltage across the 10 Ohm resistor is V = I*R = 0.59994A*10 Ohms= 5.99994V