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The ammeter shown in figure consists of ...

The ammeter shown in figure consists of a` 480(omega)`coil connected in parallel to a` 20 (omega)` shunt. Find the reading of the ammeter.

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The equivalent resistance of the ammeter is
`((480(Omega))(20(Omega)))/(480(Omega)+20(Omega))=19.2(Omega).`
the equivalent resistance of the circuit is
`140.8(Omega)+19.2(Omega)=160(Omega).`
the current is `i=(20V)/(160(Omega)=0.125A.`
this current goes through the ammeter and hence the reading of the ammeter is 0.125A.
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HC VERMA-ELECTRIC CURRENT IN CONDUCTORS-Exercises
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  7. A wire of length 1 m and radius 0.1mm gas a resistance of 100(Omega).F...

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  11. Shown a conductor of length l having a circular cross section. The rad...

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  13. Calculate the electric field in a copper wire of cross-sectional area2...

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  14. A wire has a length of 2.0m and a resistance of 5.0(Omega).Find the el...

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  15. The resistances of an iron wire and a copper wire at 20^(@)Care 3.9(Om...

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  16. The current in a conductor and the potential difference across its end...

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  17. Shown an arrangement to measure the emf (epsilon)and internal resistan...

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  18. The potential dofference between the terminals of a battery of emf6.0V...

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  19. The potential difference between the terminals of a 6.0V battery is 7....

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  20. The internal resistance of an accumulator battery of emf 6V is 10(Omeg...

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  21. Find the value of i(1)//I(2)in figure if (a)R=0.1(Omega) (b)R=1(Omega)...

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