In the circuit shown, the current in the 1 Ω resistor is

Question

In the circuit shown, the current in the 1 Ω resistor is (A) 1.3 A, from P to Q (B) 0 A (C) 0.13 A, from Q to P (D) 0.13 A, from P to Q

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/biology/fd999d5b57bee14c735326a1b83957f3-.png" /> (x+9/5)+(x-6/3)+(x/1)=0 ⇒ (3 x+27+5 x-30+15 x/15)=0 ⇒ x=(3/23) A from Q to P

Q. In the circuit shown, the current in the $1 Ω$ resistor is

1.3 A, from P to Q

0 A

0.13 A, from Q to P

0.13 A, from P to Q

$\frac{x + 9}{5} + \frac{x - 6}{3} + \frac{x}{1} = 0$
$\Rightarrow \frac{3 x + 27 + 5 x - 30 + 15 x}{15} = 0$
$\Rightarrow x = \frac{3}{23} A$
from $Q$ to $P$

Q. In the circuit shown, the current in the 1Ω resistor is

1.3 A, from P to Q

0 A

0.13 A, from Q to P

0.13 A, from P to Q

5x+9​+3x−6​+1x​=0 ⇒153x+27+5x−30+15x​=0 ⇒x=233​A from Q to P

Q. In the circuit shown, the current in the $1 Ω$ resistor is

$\frac{x + 9}{5} + \frac{x - 6}{3} + \frac{x}{1} = 0$
$\Rightarrow \frac{3 x + 27 + 5 x - 30 + 15 x}{15} = 0$
$\Rightarrow x = \frac{3}{23} A$
from $Q$ to $P$