A 50Ω resistance is connected to a battery of 5 V. A galvanometer of resistance 100Ω is to be used as an ammeter to measure current through the resistance. For this, a resistor rs is connected to the galvanometer. Which of the following connections should be employed if the measured current is within 1% of the actual current?

Current in the circuit without ammeter is Imax = V/R = 5/50 = 0.1A

Given that current in the circuit with ammeter is within 1% of the actual current, the measured current should be between 0.099A and 0.101A.

Let's analyze each option:

1. rs = 0.5Ω in series with the galvanometer:
   Total resistance of ammeter = 100Ω + 0.5Ω = 100.5Ω
   Total resistance of the circuit = 50Ω + 100.5Ω = 150.5Ω
   Measured current = V / (50Ω + 100.5Ω) = 5V / 150.5Ω ≈ 0.033A
   This is significantly less than 0.1A.

2. rs = 1Ω in parallel with the galvanometer:
   Equivalent resistance of ammeter = (100Ω * 1Ω) / (100Ω + 1Ω) ≈ 0.99Ω
   Total resistance of the circuit = 50Ω + 0.99Ω = 50.99Ω
   Measured current = 5V / 50.99Ω ≈ 0.097A
   This is within 1% of 0.1A

3. rs = 0.5Ω in parallel with the galvanometer:
   Equivalent resistance of ammeter = (100Ω * 0.5Ω) / (100Ω + 0.5Ω) ≈ 0.497Ω
   Total resistance of the circuit = 50Ω + 0.497Ω = 50.497Ω
   Measured current = 5V / 50.497Ω ≈ 0.099A
   This is within 1% of 0.1A

4. rs = 1Ω in series with the galvanometer:
   Total resistance of ammeter = 100Ω + 1Ω = 101Ω
   Total resistance of the circuit = 50Ω + 101Ω = 151Ω
   Measured current = 5V / 151Ω ≈ 0.033A
   This is significantly less than 0.1A.

Therefore, options 2 and 3 provide measured current within 1% of the actual current. Option 3 (rs = 0.5Ω in parallel with the galvanometer) gives a measured current closer to the actual value.