A.i. State Ohm's law. 10.a.ii. Name two types of electric circuits. 10.b. Calculate the maximum and minimum resistance using three resistors with values 5 Ω, 10 Ω, and 20 Ω.

Maximum resistance is obtained when the resistors are connected in series.

$$R_{max} = R_1 + R_2 + R_3$$ $$R_{max} = 5 \, \Omega + 10 \, \Omega + 20 \, \Omega$$ $$R_{max} = 35 \, \Omega$$

Minimum resistance is obtained when the resistors are connected in parallel.

$$\frac{1}{R_{min}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ $$\frac{1}{R_{min}} = \frac{1}{5 \, \Omega} + \frac{1}{10 \, \Omega} + \frac{1}{20 \, \Omega}$$ $$\frac{1}{R_{min}} = \frac{4}{20 \, \Omega} + \frac{2}{20 \, \Omega} + \frac{1}{20 \, \Omega}$$ $$\frac{1}{R_{min}} = \frac{4+2+1}{20 \, \Omega}$$ $$\frac{1}{R_{min}} = \frac{7}{20 \, \Omega}$$ $$R_{min} = \frac{20}{7} \, \Omega$$ $$R_{min} \approx 2.86 \, \Omega$$

The maximum resistance is $\boxed{\text{35 } \Omega}$ and the minimum resistance is $\boxed{\text{2.86 } \Omega}$.

10.c.i:
When Switch K1 is closed (and K2 is open).

With K1 closed and K2 open, the $5 \, \Omega$ and $3 \, \Omega$ resistors are in series.

$$R_{total} = 5 \, \Omega + 3 \, \Omega$$ $$R_{total} = 8 \, \Omega$$

The voltage source is $V = 2 \, V$.

$$I = \frac{V}{R_{total}}$$ $$I = \frac{2 \, V}{8 \, \Omega}$$ $$I = 0.25 \, A$$

The current passing through the circuit is $\boxed{\text{0.25 A}}$.

10.c.ii:
When Switches K1 and K2 are both closed.

With K1 closed and K2 closed, switch K2 short-circuits the $3 \, \Omega$ resistor. This means current will flow through K2 and bypass the $3 \, \Omega$ resistor, effectively removing it from the circuit. Only the $5 \, \Omega$ resistor remains in the path of the current.

$$R_{total} = 5 \, \Omega$$

The voltage source is $V = 2 \, V$.

$$I = \frac{V}{R_{total}}$$ $$I = \frac{2 \, V}{5 \, \Omega}$$ $$I = 0.4 \, A$$

The current passing through the circuit is $\boxed{\text{0.4 A}}$.