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To send 10% of main current through a moving coil galvanometer of resistance 99 Ω, shunt required is ______.

Asked in GUJCET 2011 · Conversion to ammeter and voltmeter

• (1) 9 Ω
• (2) 11 Ω
• (3) 10 Ω
• (4) 9.9 Ω

Step-by-step solution

Why the other options are wrong

• (1) 9 Ω is G/(n+1) with n=I/(I_g)=10. The shunt formula is S=G/(n-1): the shunt carries I-I_g, not I+I_g.
• (3) 10 Ω is the 10 of '10%' carried straight into the answer (and it is also 9.9 rounded off). The shunt is fixed by I_gG=(I-I_g)S, which gives 11 Ω.
• (4) 9.9 Ω comes from writing I_gG=IS — sending the whole current through the shunt instead of the 90% that actually goes there: 0.1×99=9.9.

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