In the diagram above, the values of V1 and V2 are respectively
1V and 1/3V
1/3V and 1V
1/3V and 2/3V
1V and 2/3V
Correct answer is D
Effective resistance in the circuit
= 3 + 2 + 1 = 6Ω
Current through the I = E/(R+V)
∴I = 2/6
=> V1 = I x R1 2/6 x 3/1 = 1V
= V2 = I x R2 = 2/6 x 2/1 = 2/3V
∴ we have 1V and 2/3V
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