Voltage-divider bias calculation: With VCC = +18 V, R1 = 4.7 kΩ (from VCC to base), and R2 = 1.5 kΩ (from base to ground), what is the nominal base bias voltage provided by the divider? Assume negligible base current loading for this calculation.

Introduction:
Voltage-divider bias is widely used because it provides a relatively stable base voltage independent of transistor β variations (within limits). Calculating the divider's nominal base voltage is a first step before accounting for base current loading and emitter feedback.

Given Data / Assumptions:

• Supply voltage VCC = 18 V.
• R1 = 4.7 kΩ from VCC to base node.
• R2 = 1.5 kΩ from base node to ground.
• Base current is negligible (ideal divider assumption).

Concept / Approach:
With negligible base loading, the base node voltage is given by the simple divider formula: V_B = VCC * (R2 / (R1 + R2)). This sets the DC bias point used by the transistor's emitter–base junction and any emitter resistor that follows.

Step-by-Step Solution:
Compute denominator: R1 + R2 = 4.7 kΩ + 1.5 kΩ = 6.2 kΩ.Form the ratio: R2 / (R1 + R2) = 1.5 / 6.2 ≈ 0.241935.Multiply by VCC: V_B = 18 V * 0.241935 ≈ 4.3548 V.Round sensibly: ≈ 4.35 V.

Verification / Alternative check:
Cross-check by direct proportionality: the base node is closer to ground because R2 is smaller than R1; a value around a quarter of VCC (18 V / 4 ≈ 4.5 V) is reasonable, confirming 4.35 V.

Why Other Options Are Wrong:
8.70 V: would require R1 = R2 (mid-supply), not the case here.2.90 V: corresponds to a different ratio; inconsistent with 1.5 kΩ and 4.7 kΩ.0.7 V: typical V_BE drop, not the divider voltage before junction drops.

Common Pitfalls:
Forgetting divider loading by base current in real circuits; accurate designs ensure divider current is 5–10 times base current to minimize shift from the ideal value.

Final Answer:
4.35 V