Question

Calculate battery current $\left(i\right)$ of the network shown in the figure.

• A. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $10\text{A}$
• B. $12\text{A}$
• C. $18\text{A}$
• D. $15\text{A}$

Answer 1

The correct option is D $15\text{A}$

If we carefully observe, we can see that all four resistors in given circuit are in parallel.

The equivalent resistance of the combination is,

$\frac{1}{R}=\frac{1}{8}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}$

$\Rightarrow R=\frac{8}{5}\Omega$

By Ohm's law current through battery is,

$I=\frac{V}{R}=\frac{24}{\frac{8}{5}}$

$\Rightarrow I=15\text{A}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (d) is the correct answer.