Question

If $R_1$ = 10 ohm, $R_2$ = 40 ohm are connected in parallel and $R_3$ = 30 ohm,$R_4$ = 20 ohm, $R_5$ = 60 ohm in another parallel combinations to a 12 V battery. What is the total resistance and current?

Answer 1

According to the question, $R_1$ and $R_2$ are connected in parallel, so their equivalent resistance s given by,

$\frac{1}{r}=\frac{1}{R_1}+\frac{1}{R_1}=\frac{1}{10}+\frac{1}{40}=\frac{4+1}{40}=\frac{5}{40}$

$r=8\Omega$.

and $R_3$, $R_4$ and $R_5$ are also connected in parallel, so their equivalent resistance s given by,

$\frac{1}{r^{\prime }}=\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}=\frac{1}{30}+\frac{1}{20}+\frac{1}{60}=\frac{2+3+1}{60}=\frac{6}{60}$

$r^{\prime }=10\Omega$

Let us assume that r and r' are connected in series, therefore the total resistance is given by,

$R=r+r=8+10=18\Omega$.

From Ohm's law we get,

Current, $I=\frac{V}{R}\frac{12}{18}=\frac{2}{3}A$.