(a)
For the correct answer: $100\ \Omega$
In a series circuit, the total resistance is the sum of individual resistances. Since the resistors are connected end-to-end in a single loop, the combined resistance is calculated as $R_{total} = R_1 + R_2 = 50\ \Omega + 50\ \Omega = 100\ \Omega$.
(b)
For the correct answer: $0.24\ \text{A}$ (which is approx. $0.25\ \text{A}$)
Using Ohm’s Law, the current $I$ is equal to the voltage $V$ divided by the total resistance $R$. With a battery voltage of $24\ \text{V}$ and a total resistance of $100\ \Omega$, the calculation is $I = 24\ \text{V} / 100\ \Omega = 0.24\ \text{A}$, which rounds to approximately $0.25\ \text{A}$.
(c)
For the correct answer: $12\ \text{V}$
The potential difference across a single resistor in series is found using $V = I \times R$. Substituting the circuit current of $0.24\ \text{A}$ and the resistance of $50\ \Omega$ gives $V_1 = 0.24\ \text{A} \times 50\ \Omega = 12\ \text{V}$, representing half the total supply voltage.
(d)
For the correct answer: $2.88\ \text{W}$
Power is the rate of energy transfer, calculated using the formula $P = I^2R$ or $P = V \times I$. Using the values for $R_1$, the power is $P = (0.24\ \text{A})^2 \times 50\ \Omega = 0.0576 \times 50 = 2.88\ \text{W}$, indicating the energy dissipated as heat per second.
(e)
For the correct answer: Resistors connected in parallel.
To minimize combined resistance, the components must be connected in parallel so the current has multiple paths. The circuit diagram should show $R_1$ and $R_2$ placed in separate branches, both connected directly across the terminals of the $24\ \text{V}$ battery.
