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Two resistors, 400 Omega, " and "800 Ome...

Two resistors, `400 Omega, " and "800 Omega` are connected in series with a 6 battery. It is desired to measure the current in the circuit. An ammeter of `10 Omega` resistance is used for this purpose. The reading of ammeter will be `N/(1210)` A. Similarly, if a voltmeter of `1000 Omega` resistance is used to measure the potential difference across the `400 Omega` resistor, the reading of voltmeter is `P/19V`. Then the value of N and P are:

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RESONANCE-CURRENT ELECTRICITY-Exercise-2 (Part-2)
  1. The current density in a cylindrical conductor of radius R varies acco...

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  2. 1 meter long metallic wire is broken into two unequal parts P and Q. ...

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  3. In the figur shown, the thermal power generated in 'Y' is maximum when...

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  4. A series parallel combination battery consisting of a large number N ...

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

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  6. A hemispherical network of radius a is made by using a conducting wire...

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  7. Find the resistance in ohm of a wire frame shaped as a cube (figure) w...

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  8. The figure is made of a uniform wire and represnts a regular five poin...

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  9. In the circuit shown in fig. E(1)=3" volt",E(2)=2" volt",E(3)=1" vole ...

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  10. The resistance of each resisor in the circuit diagram shown in figure ...

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  11. In the circuit shown in figure 16, E, F, G and H are cells of emf 2...

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  12. If the galvanometer in the circuit of figure reads zero, calculate the...

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

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  14. In the circuit shown, reading of the voltmeter connected across 400 Om...

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  15. In the given circuit the ammeter A(1)" and "A(2) are ideal and the amm...

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  16. Two resistors, 400 Omega, " and "800 Omega are connected in series wit...

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