Question

Two resistors of $2\Omega$ and $4\Omega$ are connected in parallel. Two more resistors $3\Omega$ and $6\Omega$ are also connected in parallel. These two combinations are in series with a battery of emf $5$V and internal resistance $0.7\Omega$. Calculate the current through $6\Omega$ resistor.

Answer 1

Given that $2\Omega$ and $4\Omega$ are connected in parallel,

So, $\frac{1}{2}+\frac{1}{4}=\frac{1}{R_{eq1}}$

$\Rightarrow \frac{1}{R_{eq1}}=\frac{6}{8}=\frac{3}{4}$

$\Rightarrow \frac{4}{3}=1.3333\Omega$

Again $3\Omega$ and $6\Omega$ are coonected in parallel,

So, $\frac{1}{3}+\frac{1}{6}=\frac{1}{R_{eq2}}$

$\Rightarrow \frac{1}{R_{eq2}}=\frac{9}{18}$

$\Rightarrow \frac{9}{18}=2\Omega$

Now, $R_{eq1}$ and $R_{eq2}$ are coonected in series,

So, $R_{eq}=R_{eq1}+R_{eq2}+InternalResistance$

$\Rightarrow R_{eq}=1.333+2+0.7$

$\Rightarrow R_{eq}=4.0333\Omega$

So,

$I=\frac{V}{R}=\frac{5}{4.033}=1.239Amp\simeq 1.24Amp$

$I_4=\frac{R_3\cdot I}{R_3+R_2}$

$\Rightarrow I=\frac{3\times 1.24}{3+6}=\frac{3.72}{9}=0.4133Amp$

So, The current through $6\Omega$ is $0.4133Amp$