A student is provided with a variable voltage source V, a test resistor R_T=10 Ω, two identical galvanometers G₁ and G₂ and two additional resistors, R₁=10 M Ω and R₂=0.001 Ω. For conducting an experiment to verify ohm's law, the most suitable circuit is:

Solution

Physics sir · 3 steps

Make one galvanometer a voltmeter and the other an ammeter, then check which circuit wires each correctly.

1. Voltmeter. A voltmeter goes in parallel with R_T and must draw almost no current. So the galvanometer needs a large resistance in series, which here is R₁ = 10 MΩ.
2. Ammeter. An ammeter goes in series with R_T and must add almost no resistance. So the galvanometer needs a small shunt in parallel, which here is R₂ = 0.001 Ω.

   Option	G₁ (across R_T)	G₂ (in the main line)	Works?
   (a)	shunted by R₂: low resistance, shorts R_T	R₁ in parallel: does almost nothing	No
   (b)	shunted by R₂: shorts R_T	R₁ in series: 10 MΩ blocks the current	No
   (c)	R₁ in series: voltmeter	R₂ as shunt: ammeter	Yes
   (d)	R₁ in parallel: still a bare galvanometer, not a voltmeter	R₂ in series: still a bare galvanometer, not an ammeter	No

3. So. Only circuit (c) has G₁ with R₁ in series across R_T and G₂ shunted by R₂ in series with R_T. The answer is (c).

The idea to keep. Voltmeter: galvanometer plus a big series resistance, connected in parallel. Ammeter: galvanometer plus a tiny shunt, connected in series. Big resistors go in series and tiny ones in parallel, never the other way round.

Answer from the NTA final answer key. Solution by PrepWiser.