A current of 6A enters one corner P of an equilateral triangle PQR having 3 wires of resistances 2- each and leaves by the corner R- Then the currents I1 and I2 are-

The correct option is B 4A, 2A By using Kirchhoff's current law at P: #160; I_1+I_2=6 .......(i)By Kirchhoff's voltage law to the closed circuit PQR ...

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Standard XII

Physics

Mixed Combination of Cells

A current of ...

Question

A current of $6$ A enters one corner P of an equilateral triangle PQR having $3$ wires of resistances $2 Ω$ each and leaves by the corner R. Then the currents $I_{1}$ and $I_{2}$ are.

2 A, 4 A

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4 A, 2 A

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1 A, 2 A

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2 A, 3 A

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Solution

The correct option is B $4 A, 2 A$
By using Kirchhoff's current law at P:
$I_{1} + I_{2} = 6$ .......(i)
By Kirchhoff's voltage law to the closed circuit PQRP:
$- 2 I_{1} - 2 I_{1} + 2 I_{2} = 0$
$\Rightarrow - 4 I 1 + 2 I 2 = 0$
$\Rightarrow 2 I_{1} - I_{2} = 0$ .......(ii)
On adding Equations (i) and (ii), we get,
$I_{1} + I_{2} + 2 I_{1} - I_{2} = 6 + 0$
$\Rightarrow 3 I_{1} = 6$
$\Rightarrow I_{1} = 2 A$
Now, on putting the value of $I_{1}$ in equation (i):
$I_{2} = 6 - 2 = 4 A$
Hence, the correct option is $(A)$

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A current of 6A enters one corner P of an equilateral triangle PQR having 3 wires of resistances 2Ω each and leaves by the corner R. Then the currents I1 and I2 are.

A current of $6$ A enters one corner P of an equilateral triangle PQR having $3$ wires of resistances $2 Ω$ each and leaves by the corner R. Then the currents $I_{1}$ and $I_{2}$ are.