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PhysicsNEET 2024Cells, EMF, internal resistance, combinations

Current Electricity

The terminal voltage of a battery whose emf is 10 V and internal resistance 1 Ω1\ \Omega, when connected through an external resistance of 4 Ω4\ \Omega, is:

  1. A.6 V
  2. B.8 V✓
  3. C.10 V
  4. D.4 V

Solution

Current I=εR+r=104+1=2I = \frac{\varepsilon}{R + r} = \frac{10}{4+1} = 2 A. Terminal voltage V=ε−Ir=10−2=8V = \varepsilon - Ir = 10 - 2 = 8 V.

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