Two cells A and B of emf 2V and 1.5 V respectively, are connected

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441.

Assume that each diode as shown in the figure has a forward bias resistance of 50 Ω and an infinite reverse bias resistance. The current through the resistance 150 Ω is

          

  • 0.66 A

  • 0.05 A

  • Zero

  • 0.04 A


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442.

Two cells A and B of emf 2V and 1.5 V respectively, are connected as shown in figure through an external resistance 10 Ω. The internal resistance of each cell is 5 Ω. The potential difference EA and EB across the terminals of the cells A and B respectively are

              

  • EA = 2.0 V, EB = 1.5 V

  • EA = 2.125 V, EB  = 1.375 V

  • EA = 1.875 V, EB = 1.625 V

  • EA = 1.875 V, EB = 1.375 V


C.

EA = 1.875 V, EB = 1.625 V

The figure can be redrawn as

        

The current through the circuit

        i = net emfeffective resistance  = 2 - 1.55 + 5 + 10 = 0.520 = 140  = 0.025 A

The terminal potential difference of the batteries

         VA = εA − irA = 2 − 0.025 × 5

              = 2 − 0.0125 = 1.875 V

and    VB = εB + irB = 1.5 + 0.025 × 5

              = 1.5 + 0.0125

              = 1.625 V


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