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The circuit is reduced to find equivalent resistance as follows;


In (1), 2Ω, 4 Ω, and 2 Ω are in series. Their equivalent resistance is 8 Ω. In (2), the two 8 Ω resistors are in parallel and their equivalent resistance is 4 Ω. In (3) 2 Ω, 4 Ω, and 2 Ω are in series. Their equivalent resistance is 8 Ω which is in parallel with the 8 Ω resistance as shown in (4). The total resistance in the circuit
R= 3+4+2+1( internal resistance)=10 Ω. The current through the 3 Ω resistor
I = E/R = 10/10
I = 1A