The magnitude and direction of the current in the circuit shown will be

Question

The magnitude and direction of the current in the circuit shown will be (A) 7/3 A from a to b through e (B) 7/3 A from b to a through e (C) 1A from b to a through e (D) 1A from a to b through e

Tardigrade Admin · Accepted Answer

Since E1 (10 V) > E2 (4V) So, current in the circuit will be clockwise, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/10c1b70dab06829c210e8166a5b85225-.png" /> Applying Kirchhoffâs voltage law -1 Ã i - 10 - 4 Ã -2 Ã i -3i = 0 = 1A (a to b via e) Current I=(V/R)=(10-4/6)=1 A

Q. The magnitude and direction of the current in the circuit shown will be

$7/3$ A from a to b through $e$

$7/3$ A from b to a through $e$

$1 A$ from b to a through $e$

$1 A$ from a to b through $e$

Since $E_{1} (10 V) > E_{2} (4 V)$
So, current in the circuit will be clockwise,

Applying Kirchhoff’s voltage law
$- 1 \times i - 10 - 4 \times - 2 \times i - 3 i = 0$
$= 1 A$ (a to b via e)
Current $I = \frac{V}{R} = \frac{10 - 4}{6} = 1 A$

Q. The magnitude and direction of the current in the circuit shown will be

7/3 A from a to b through e

7/3 A from b to a through e

1A from b to a through e

1A from a to b through e

Since E1​(10V)>E2​(4V) So, current in the circuit will be clockwise, Applying Kirchhoff’s voltage law −1×i−10−4×−2×i−3i=0 =1A (a to b via e) Current I=RV​=610−4​=1A

$7/3$ A from a to b through $e$

$7/3$ A from b to a through $e$

$1 A$ from b to a through $e$

$1 A$ from a to b through $e$

Since $E_{1} (10 V) > E_{2} (4 V)$
So, current in the circuit will be clockwise,

Applying Kirchhoff’s voltage law
$- 1 \times i - 10 - 4 \times - 2 \times i - 3 i = 0$
$= 1 A$ (a to b via e)
Current $I = \frac{V}{R} = \frac{10 - 4}{6} = 1 A$