Two identical moving coil galvanometer have 10 Ω resistance and full scale deflection at 2   μ A current. One of them is converted into a voltmeter of 100 m V full scale reading and the other into an Ammeter of 1 m A full scale current using appropriate resistors. These are then used to measure the voltage and current in the Ohm's law experiment with R = 1000 Ω resistor by using an ideal cell

Step-by-step solution

Voltmeter rom galvanometer arrangement. Given: R g = 10 Ω and I g = 2 μ A Required voltage range, V = 100 × 10 - 3 v o l t s ⇒ V = I g R g + R v = 10 - 1 ⇒ 10 - 1 2 × 10 - 6 = R g + R v a s , I g = 2 μ A ⇒ 5 × 10 4 Ω ≈ R v ( R v 10 5 Ω ) ⇒ Option ( A ) is incorrect. Galvanometer to Ammeter arrangement Δ V G = Δ V S I g ⋅ R g = I - I g ⋅ S ⇒ S = I g ⋅ R g I - I g = 2 × 10 - 6 × 10 10 - 3 - 2 × 10 - 6 ⇒ S = 2 × 10 - 5 × 10 3 = 2 × 10 - 2 ⇒ S = 20 m Ω ⇒ ( C ) is correct. Circuit diagram for ohm's law Resistance ammete