Physics · Moving Charges And Magnetism · NEET
An ammeter goes in series with the circuit, so the FULL current I must reach it. But the galvanometer coil can only take a tiny current I_g (like 1 mA) before it burns or goes off-scale. A parallel shunt gives the extra current (I - I_g) a low-resistance side path, so only I_g goes through the coil. In series everything would be forced through the coil and destroy it.
Very small, usually a fraction of an ohm. It must be much smaller than the galvanometer resistance R_g so that most of the current takes the shunt path. Since R_s = I_g R_g / (I - I_g) and (I - I_g) is large while I_g is tiny, the answer comes out very small (e.g. 0.01 ohm).
An ammeter is placed in series, so it becomes part of the circuit. If its resistance were high, it would reduce the very current it is trying to measure. Adding a small shunt in parallel with the galvanometer drops the combined resistance to nearly zero, so the ammeter does not disturb the circuit. An ideal ammeter has zero resistance.
Always just I_g, the full-scale deflection current, no matter how large the range I is. That is the whole point: at full range the pointer reads maximum because I_g flows through the coil, while the remaining (I - I_g) quietly passes through the shunt. The scale is then re-marked from 0 to I.
It is R_g and R_s in parallel: R_A = R_g R_s / (R_g + R_s). Because R_s is very small, R_A is even smaller than R_s, which is exactly what we want for a good ammeter.
A galvanometer of resistance 100 ohm gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0 - 10 A. The shunt required is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
A shunt is a small resistance connected in parallel with the galvanometer coil. It provides a low-resistance path for most of the current so that only the safe full-scale current I_g flows through the delicate coil.
R_s = I_g R_g / (I - I_g), where I_g is the galvanometer full-scale current, R_g is its resistance, and I is the maximum current the ammeter should read.
Yes. To read a larger current I, you need a smaller shunt. As I increases, (I - I_g) increases, so R_s = I_g R_g / (I - I_g) becomes smaller. A smaller shunt lets more current bypass the coil.
Because it is connected in series, any resistance it adds would reduce the circuit current and give a wrong reading. The small shunt makes the ammeter's effective resistance nearly zero, so it does not disturb the current it measures.
For an ammeter you add a SMALL shunt in PARALLEL to lower resistance. For a voltmeter you add a LARGE resistance in SERIES to raise resistance, so it draws almost no current when placed across two points.
