PhysicsIn the circuit shown in figure potential difference between points A and B is 16 V . The current passing through 2 Ω resistance will be

In the circuit shown in figure potential difference between points A and B is $16 V$ . The current passing through $2 Ω$ resistance will be

$V_{A} - V_{B} = 16 V$

$∴ 4 i_{1} + 2 (i_{1} + i_{2}) - 3 + 4 i_{1} = 16 V .......... (1)$

Using Kirchhoff’s second law in the closed loop we have

$9 - i_{2} - 2 (i_{1} + i_{2}) = 0 .............. (2)$

Solving Eqs. (1) and (2), we get

$i_{1} = 1.5 A$ and $i_{2} = 2 A$

Current through $2 Ω$ resistor

$= 2 + 1.5 = 3.5 A$

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