Solutions
Concept:
The BJT configuration for a pnp and npn transistors are as shown:


The emitter-collector voltage can be calculated using KVL
Analysis:
The configuration given is of a PNP transistor.
Given, VE = 3 V

VEB = 600 × 10-3 V = 0.6 V
VE – VB ⇒ 0.6 V
3 – VB ⇒ 0.6 V
So, VB = 2.4 Volts
Applying Ohm’s law across 60 kΩ resistor,
IC = β IB (where β is the current gain)
Given, β = 50,
IC = 50 × 0.04 × 10-3 A = 2 mA
IC = 2 × 10-3 A
Applying Ohms law across 500 Ω resistor, we get:
Now,
VEC = VE – VC = 3 – 1 = 2 Volts
So, VEC = 2 Volts
Common Mistakes:
Mark emitter & collector terminals accordingly. It is common to notice that students always apply npn-transistor concepts even for a given pnp transistor.