To solve this problem, we can apply Ohm's Law and the concept of proportional voltage division in a series circuit.

Steps:

1. Calculate the total resistance of the circuit: Since the resistors are in series, the total resistance $R_{\text{total}}$ is simply the sum of the individual resistances: $R_{\text{total}} = R_A + R_B = 30\, \Omega + 10\, \Omega = 40\, \Omega$

2. Determine the total current in the circuit: We use Ohm's Law $V = I \cdot R$ to calculate the total current $I$. Rearranging the formula to solve for current: $I = \frac{V_{\text{total}}}{R_{\text{total}}}$ where $V_{\text{total}} = 122 \, \text{V}$ is the applied voltage and $R_{\text{total}} = 40 \, \Omega$. $I = \frac{122\, \text{V}}{40\, \Omega} = 3.05\, \text{A}$

3. Calculate the voltage drop across resistor A: Using Ohm's Law again for resistor A: $V_A = I \cdot R_A$ where $I = 3.05\, \text{A}$ and $R_A = 30\, \Omega$: $V_A = 3.05\, \text{A} \times 30\, \Omega = 91.5\, \text{V}$

Thus, the voltage drop across resistor A is 91.5 V.

Final Answer: 91.5 V

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Do you need any further details or have additional questions?

Here are 5 related questions for further practice:

1. What is the voltage drop across resistor B?
2. How would the voltage drop change if the resistances were reversed?
3. If the total resistance were 60 Ω instead of 40 Ω, how would the voltage drop across A change?
4. What is the current if the applied voltage were 150 V?
5. How does the total resistance affect the overall current in the series circuit?

Tip: In a series circuit, the current is the same across all components, but the voltage drop is divided based on the resistance of each component.