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NEETPhysicsCurrent Electricity

Two batteries and two resistors are connected in series to form a single-loop circuit. The first battery has an unknown EMF E and the second battery has an EMF of 4 V . The batteries are connected in opposition. The two resistors have resistances of 2 and 3 . If a steady current of 2 A flows through the circuit such that it leaves the positive terminal of the first battery, what is the value of the unknown EMF E ? (A

Options

  1. A14 V
  2. B10 V
  3. C8 V
  4. D6 V

Correct answer

A. 14 V

Step-by-step solution

Since the batteries are connected in opposition and the current leaves the positive terminal of the first battery, the first battery dominates. The net EMF in the circuit is E_ net = E - 4 . The total resistance of the circuit is the sum of the series resistors: R_ net = 2 + 3 = 5 . According to Ohm's law for the complete circuit, the current is: I = E_ net R_ net Substituting the given values: 2 = E - 4 5 10 = E - 4 E = 14 V . Answer: 14 V

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