In the figure shown, the current in the
Question:

In the figure shown, the current in the $10 \mathrm{~V}$ battery is close to:

  1. $0.36$ A from negative to positive terminal.

  2. $0.71$ A from positive to negative terminal.

  3. $0.21$ A from positive to negative terminal.

  4. $0.42$ A from positive to negative terminal.


Correct Option: , 3

Solution:

$\mathrm{E}_{\mathrm{eq}}=\frac{20 \times 10}{17}=\frac{200}{17}$

and $\mathrm{R}_{\text {eq }}=\frac{7 \times 10}{17}=\frac{70}{17}$

$\therefore \quad I=\frac{\frac{20}{17}-10}{4+\frac{70}{17}}=0.21 \mathrm{~A}$

from +ve to -ve terminal

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