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) 1 A from b to a through e (D) 1 A from a to b through e

Tardigrade Admin · Accepted Answer

Since E1(10 V ) > E2(4 V ) So current in the circuit will be clockwise. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/15c8351e8542822e609a718df0fd72f4-.png" /> Applying Kirchhoff's voltage law, -1 × i+10-4-2 × i-3 i=0 ⇒ i=1 A (a to b via e) ∴ Current =(V/R)=(10-4/6)=1.0 ampere

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

$\frac{7}{3} A$ from $a$ to $b$ through $e$

$\frac{7}{3} A$ 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,
$- 1 \times i + 10 - 4 - 2 \times i - 3 i = 0$
$\Rightarrow i = 1 A (a$ to $b$ via $e)$
$∴$ Current $= \frac{V}{R} = \frac{10 - 4}{6} = 1.0$ ampere

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

37​A from a to b through e

37​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 ⇒i=1A(a to b via e) ∴ Current =RV​=610−4​=1.0 ampere

$\frac{7}{3} A$ from $a$ to $b$ through $e$

$\frac{7}{3} A$ 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,
$- 1 \times i + 10 - 4 - 2 \times i - 3 i = 0$
$\Rightarrow i = 1 A (a$ to $b$ via $e)$
$∴$ Current $= \frac{V}{R} = \frac{10 - 4}{6} = 1.0$ ampere