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In the circuit shown below E₁ = 4.0 V, $R_{1} = 2 Ω$ E₂ = 6.0 V, $R_{2} = 4 Ω$ and $R_{3} = 2 Ω .$ The current I₁ is:

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Correct option is B

For loop ABEFA−2I2−I1−4I2+6=0 …(1)2I1−6I2+6=0 or I1=3I2−1 …(1)For loop BCDEB4−2I1+2I2−I1=0I2=2I1−2…(2)Put in (1)I1=32I1−2−1I1=6I1−9 OR I1=1.8A

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In the circuit shown below E1 = 4.0 V, R1=2Ω E2 = 6.0 V, R2=4 Ω and R3=2 Ω. The current I1 is:

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In the circuit shown below E₁ = 4.0 V, $R_{1} = 2 Ω$ E₂ = 6.0 V, $R_{2} = 4 Ω$ and $R_{3} = 2 Ω .$ The current I₁ is: