Type all the answers

Step 1: Calculate the equivalent resistance of $R_2$ and $R_3$ which are connected in parallel. The formula for two resistors in parallel is: $\frac{1}{R_{p}} = \frac{1}{R_2} + \frac{1}{R_3}$ Substitute the given values $R_2 = 20 \, \Omega$ and $R_3 = 30 \, \Omega$: $\frac{1}{R_{p}} = \frac{1}{20 \, \Omega} + \frac{1}{30 \, \Omega}$ Find a common denominator, which is $60$: $\frac{1}{R_{p}} = \frac{3}{60 \, \Omega} + \frac{2}{60 \, \Omega}$ $\frac{1}{R_{p}} = \frac{3+2}{60 \, \Omega} = \frac{5}{60 \, \Omega}$ Now, invert the fraction to find $R_p$: $R_{p} = \frac{60 \, \Omega}{5} = 12 \, \Omega$

Step 2: Calculate the total resistance of the circuit. The parallel combination $R_p$ is in series with $R_1$. The formula for resistors in series is: $R_{total} = R_1 + R_{p}$ Substitute the values $R_1 = 10 \, \Omega$ and $R_p = 12 \, \Omega$: $R_{total} = 10 \, \Omega + 12 \, \Omega$ $R_{total} = 22 \, \Omega$

The total resistance of the circuit is $\boxed{22 \Omega}$.