In the figure shown, the current in the 10 V battery is close to:

Question

In the figure shown, the current in the 10 V battery is close to: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/6be9a70ae6e2566e47aea4977c23c0c1-.png" /> (A) 0.36 A from negative to positive terminal. (B) 0.71 A from positive to negative terminal. (C) 0.21 A from positive to negative terminal. (D) 0.42 A from positive to negative terminal

Tardigrade Admin · Accepted Answer

E eq =(20 × 10/17)=(200/17) and R eq =(7 × 10/17)=(70/17)

∴ I=((20/17)-10/4+(70)17)=0.21 A from +ve to -ve terminal

Q. In the figure shown, the current in the $10 V$ battery is close to :

$0.36 A$ from negative to positive terminal.

$0.71 A$ from positive to negative terminal.

$0.21 A$ from positive to negative terminal.

$0.42 A$ from positive to negative terminal

$E_{e q} = \frac{20 \times 10}{17} = \frac{200}{17}$
and $R_{e q} = \frac{7 \times 10}{17} = \frac{70}{17}$

$∴ I = \frac{\frac{20}{17} - 10}{4 + \frac{70}{17}} = 0.21 A$
from +ve to -ve terminal

Q. In the figure shown, the current in the 10V battery is close to :

0.36A from negative to positive terminal.

0.71A from positive to negative terminal.

0.21A from positive to negative terminal.

0.42A from positive to negative terminal