Let's solve this step by step for each circuit using Ohm's Law and the power formula.

(i) Circuit:

1. Ohm’s Law: $V = IR$, where:

   • $V = 10 \, \text{V}$ (voltage of the battery),
   • $R = 50 \, \Omega$ (resistance of the resistor).

   Solve for $I$: $I = \frac{V}{R} = \frac{10}{50} = 0.2 \, \text{A}.$

   Thus, the current $I = 0.2 \, \text{A}$.

2. Power Dissipated: Power ($P$) is given by: $P = V \cdot I = 10 \cdot 0.2 = 2 \, \text{W}.$

   Alternatively, using $P = I^2 R$: $P = (0.2)^2 \cdot 50 = 0.04 \cdot 50 = 2 \, \text{W}.$

(ii) Circuit:

1. Ohm’s Law: Here, the current $I = 2 \, \text{A}$, and the resistance $R = 50 \, \Omega$.

   Solve for $V$: $V = IR = 2 \cdot 50 = 100 \, \text{V}.$

   Thus, the voltage $V = 100 \, \text{V}$.

2. Power Dissipated: Power ($P$) is given by: $P = V \cdot I = 100 \cdot 2 = 200 \, \text{W}.$

   Alternatively, using $P = I^2 R$: $P = (2)^2 \cdot 50 = 4 \cdot 50 = 200 \, \text{W}.$

Final Results:

1. Circuit (i):

   • $I = 0.2 \, \text{A}$,
   • $P = 2 \, \text{W}$.

2. Circuit (ii):

   • $V = 100 \, \text{V}$,
   • $P = 200 \, \text{W}$.

Would you like more explanation or have further questions?

5 Questions to Explore Further:

1. How does Ohm's Law change in parallel circuits?
2. What happens to the current if the resistance is doubled in circuit (i)?
3. How is power dissipation related to energy transfer?
4. Can you calculate the total energy consumed in 1 hour for circuit (ii)?
5. What would happen if the resistor in circuit (ii) were replaced with a 25 $\Omega$ resistor?

Tip:

Always double-check units for consistency when solving electrical circuit problems!