Question

Find the current through the $10\Omega$ resistance in network shown below Choose answer:

• A. ${\frac{5}{8}}_A$
• B. ${\frac{10}{7}}_A$
• C. ${\frac{20}{17}}_A$
• D. ${\frac{5}{7}}_A$

Answer 1

The correct option is C ${\frac{20}{17}}_A$

Let us replace the two cells and $4\Omega$ and $2\Omega$ resistances with a single battery of Emf E and resistance with R.

$\Rightarrow E=\frac{\frac{e_1}{R_1}+\frac{e_2}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}}=\frac{\frac{20}{4}+\frac{10}{2}}{\frac{1}{4}+\frac{1}{2}}=\frac{10\times 4}{3}=\frac{40}{3}V$

$\Rightarrow R={(\frac{1}{R_1}+\frac{1}{R_2})}^{-1}={(\frac{1}{4}+\frac{1}{2})}^{-1}=\frac{4}{3}\Omega$

$\Rightarrow R_{eq}=\frac{4}{3}+10=\frac{34}{3}\Omega$

$\Rightarrow I=\frac{40}{34}=\frac{20}{17}A$

Therefore option C is correct.