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Voltage divider calculation — an ideal 10 V DC source feeds a series divider of R1 = 1 kΩ and R2 = 9 kΩ. What is the voltage drop across R1 (that is, VR1)?

Difficulty: Easy

1 V
9 V
10 V
19 V
0.5 V

Correct Answer: 1 V

Explanation:

Introduction / Context:
Voltage dividers are used to generate a fraction of a supply voltage. The share of the source that appears across each series resistor is set by its resistance relative to the total. This quick exercise reinforces the voltage-division rule for a simple two-resistor chain.

Given Data / Assumptions:

Ideal source: V_S = 10 V.
Series resistors: R1 = 1 kΩ, R2 = 9 kΩ.
No loading on the divider output.

Concept / Approach:
In a series chain with current I, the drop across each resistor is V_i = I * R_i. The common current is I = V_S / (R1 + R2). Therefore, V_R1 = I * R1 = V_S * (R1 / (R1 + R2)). This is the standard voltage-divider expression and is widely used in biasing and sensing circuits.

Step-by-Step Solution:

Compute the total resistance: R_T = 1 kΩ + 9 kΩ = 10 kΩ.Find the current: I = V_S / R_T = 10 V / 10 kΩ = 1 mA.Compute VR1: VR1 = I * R1 = 1 mA * 1 kΩ = 1 V.Alternatively use divider ratio: VR1 = 10 V * (1 kΩ / 10 kΩ) = 1 V.

Verification / Alternative check:
Check sum of drops: VR1 + VR2 = 1 V + 9 V = 10 V, which equals the source voltage as required by KVL.

Why Other Options Are Wrong:

9 V and 10 V: correspond to the drop on R2 or the full source, not VR1.19 V: exceeds the source; impossible without an active element.0.5 V: would require a different ratio (R1 half of 1 kΩ relative to total).

Common Pitfalls:
Forgetting units (kΩ vs Ω); mixing up which resistor is being measured; ignoring divider loading when present in real designs.

Voltage divider calculation — an ideal 10 V DC source feeds a series divider of R1 = 1 kΩ and R2 = 9 kΩ. What is the voltage drop across R1 (that is, VR1)?

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