Question

Resistors $R_1$=10$\Omega$, $R_2$ = 40$\Omega$, $R_3$ = 30$\Omega$, $R_4$ = 20$\Omega$, $R_5$ = 60$\Omega$ and a 12V battery is connected as shown.

Calculate:
(a) the total resistance
(b) the total current flowing in the circuit.

Answer 1

The situation is shown in (figure)



For $R_1$ and $R_2$ in parallel

$\frac{1}{R_{p1}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{10}+\frac{1}{40}=\frac{5}{40}=\frac{1}{8}$

or $R_{p1}=8\Omega$

For $R_3,R_4$ and $R_5$ is parallel

$R_{p2}=10\Omega$

(a) For $R_{p1}$ and $R_{p2}$ in series.

Total resistance, $R=R_{p1}+R_{p2}$

Putting values, we get, R = 8 + 10 = 18

Total resistance, R = 18 $\Omega$.

(b) From relation, (Ohm's law) $R=\frac{V}{I}$

We have, $I=\frac{V}{R}$

Putting values, we get, I = 0.67

Total current, I = 0.67 A