Question 33: The magnitude and direction of the current in the following circuit is
(1) \frac{5}{9} A from A to B through E
(2) 1.5 A from B to A through E
(3) 0.2 A from B to A through E
(4) 0.5 A from A to B through E
Answer: Option (4)
Explanation:
In the given closed circuit, all resistances are connected in series.
Hence, the total resistance of the circuit is the sum of individual resistances.
The resistances are 2 \Omega, 1 \Omega,
and 7 \Omega. Therefore, the total resistance is R = 2 + 1 + 7 = 10 \Omega.
There are two cells of emf 10 \mathrm{~V} and 5 \mathrm{~V}
connected in opposition, as indicated by their polarities in the circuit.
Hence, the net emf of the circuit is E = 10 - 5 = 5 \mathrm{~V}.
Using Ohm’s law, the current in the circuit is given by I = \frac{E}{R}.
Substituting the values, I = \frac{5}{10}.
Thus, the current is 0.5 \mathrm{~A}.
The direction of current is from the positive terminal of the stronger cell 10 \mathrm{~V},
which corresponds to current flowing from A to B through point E.
Hence, the correct answer is Option (4).