The circuit shown consists of a 5V DC power source and two resistors, $R_1 = 10 \, \text{k}\Omega$ and $R_2 = 100 \, \text{k}\Omega$, connected in series. Let's calculate the quantities step by step.

1. Total Equivalent Resistance ($R_{\text{eq}}$):

For resistors in series, the equivalent resistance is the sum of the individual resistances: $R_{\text{eq}} = R_1 + R_2 = 10 \, \text{k}\Omega + 100 \, \text{k}\Omega = 110 \, \text{k}\Omega$

2. Current $I_1$:

Using Ohm's Law $I = \frac{V}{R}$, the current $I_1$ through the series circuit is: $I_1 = \frac{V_{\text{total}}}{R_{\text{eq}}} = \frac{5 \, \text{V}}{110 \, \text{k}\Omega} = \frac{5}{110 \times 10^3} \, \text{A} \approx 45.45 \, \mu \text{A}$

3. Voltage Drop Across $R_1$:

The voltage drop across $R_1$ is given by Ohm's Law: $V_{R_1} = I_1 \times R_1 = 45.45 \, \mu \text{A} \times 10 \, \text{k}\Omega = 45.45 \times 10^{-6} \, \text{A} \times 10 \times 10^3 \, \Omega = 0.4545 \, \text{V}$

4. Voltage Drop Across $R_2$:

Similarly, the voltage drop across $R_2$ is: $V_{R_2} = I_1 \times R_2 = 45.45 \, \mu \text{A} \times 100 \, \text{k}\Omega = 45.45 \times 10^{-6} \, \text{A} \times 100 \times 10^3 \, \Omega = 4.545 \, \text{V}$

Summary of Results:

• Current $I_1 = 45.45 \, \mu \text{A}$
• Voltage drop across $R_1 = 0.4545 \, \text{V}$
• Voltage drop across $R_2 = 4.545 \, \text{V}$

Let me know if you'd like more details or have any questions!

Here are 5 related questions to deepen your understanding:

1. What happens to the current if we reduce $R_1$ to 5 kΩ?
2. How would the voltage drops change if the total voltage increased to 10V?
3. What would the current be if $R_2$ were removed from the circuit?
4. How does increasing the resistance of $R_2$ affect the current in the circuit?
5. Can you calculate the power dissipated by $R_1$ and $R_2$?

Tip: In a series circuit, the current is the same through all components, but the voltage drops across each resistor depend on its resistance.