Question

If in figure $R_1=10\Omega$, $R_2=40\Omega$, $R_3=30\Omega$, $R_4=20\Omega$, $R_5=60\Omega$, and a 12 V battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.

Answer 1

(a) 10 ohm and 40 ohm are in parallel, so their equivalent resistance is

$\frac{1}{R_{eq1}}=\frac{1}{10}+\frac{1}{40}$

Gives, $R_{eq1}=8\Omega$

30ohm, 20ohm, 60ohm are in parallel

$\therefore \frac{1}{R_{eq2}}=\frac{1}{30}+\frac{1}{20}+\frac{1}{60}=\frac{2+3+1}{60}=\frac{1}{10}$

$\therefore R_{eq2}=10\Omega$

Now $R_{eq1}$ and $R_{eq2}$ are in series.

Total resistance of the circuit $R=R_{eq1}+R_{eq2}=8+10=18\Omega$

(b) Total current=$\frac{V}{R}=\frac{12}{18}=0.67A$