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Siix lead-acid type of secondary cells, ...

Siix lead-acid type of secondary cells, each of emf 2.0 V and internal resistancce `0.015 Omega` are joined in series to provide a supply to a resistance of `8.5Omega`. Determine : (i) the current draw from the fupply and (ii) its terminal voltage.

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The correct Answer is:
(i) `112/8.59=1.4 A.`
(ii) `(12xx8.5)/(8.59)=11.9V`
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RESONANCE ENGLISH-CURRENT ELECTRICITY-Exercise-1 (Part-1)
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  5. Suppose you have three resistor of (20 Omega),(50 Omega)and (100 Omega...

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  6. Three bulbs,each having a resistance of 180(Omega)are connected in par...

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  7. Consider the circuit shown in figure.Find the current through the10(Om...

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  8. Siix lead-acid type of secondary cells, each of emf 2.0 V and internal...

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  9. In the figure each cell has an emf of 1.5 V and internal resistnce of ...

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  10. In the circuit shown all live resistors have the same value 200 ohms a...

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  11. Find the currents through the three resistors shown in Figure

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

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  13. A galvenometer has a resistance of 30 Omega, and a current of 2 mA is ...

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  14. A voltmeter of resistances 400(Omega)is used to measure the potential ...

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  15. An electrical circuit is shown in figure. Calculate the potential diff...

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  16. A battery of emf 1.4 V and internal resistance 2 Omega is connected to...

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  17. A potentiometer wire AB having length L and resistance 12 r is joined ...

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  18. Figure 6.13 shows a 2.0 V potentiometer used for the determination of ...

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  19. Figure 6.36 shows a potentiometer with a cell of emf 2.0 V and interna...

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  20. Figure 6.32 shows a meter bridge in the (which is nothing but a partic...

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