The first junction on transistor consisting of two back to back P-N junction was invented in 1951 by William Shockley. Later new Bipolar Junction Transistor (BJT) were invented. Transistor has three doped regions forming two P-N junction between them. There are two types of transistors (i) P-N-P and (ii) N-P-N. Each transistor has three segments (i) Emitter (ii) Base (iii) Collector. The size and level of doping is different of all three segments. For connecting in the circuit three terminals of a transistor are available - emitter, base and collector and it can be connected in three configuration Common Emitter (CE), Common Base (CB) and Common Collector (CC). The transistor can be used as a device application depending on configuration used, the biasing done to the segments and operate in the region namely cut off, Active and Saturation. It can work as an amplifier, a switch and an oscillator too. Based on transistor knowledge give answer.
A silicon transistor having dc current amplification of \( \beta = 100 \) is used in a CE Configuration in a circuit below. In this circuit \( I_B = 40\,\mu\mathrm{A}\;(40 \times 10^{-6}\,\mathrm{A}) \). When the load resistance of \( 2\,\mathrm{k\Omega} \) is connected to the collector with \( V_{CC} = 10\,\mathrm{V} \). Select the correct option giving value for collector to emitter Voltage \( V_{CE} \).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2\,\mathrm{V} \)
In the CE configuration the collector current is fixed by the dc current gain, \( I_C = \beta I_B \). The supply voltage \( V_{CC} \) is shared between the load resistance and the transistor, so by Kirchhoff's loop rule in the collector-emitter circuit \( V_{CE} = V_{CC} - I_C R_L \).
\( \beta = 100 \)
\( I_B = 40\,\mu\mathrm{A} = 40\times10^{-6}\,\mathrm{A} \)
\( R_L = 2\,\mathrm{k\Omega} = 2\times10^{3}\,\Omega \)
\( \displaystyle I_C = \beta I_B = 100 \times 40\times10^{-6} = 4\times10^{-3}\,\mathrm{A} = 4\,\mathrm{mA} \)
\( \displaystyle I_C R_L = 4\times10^{-3}\,\mathrm{A} \times 2\times10^{3}\,\Omega = 8\,\mathrm{V} \)
\( \displaystyle V_{CE} = V_{CC} - I_C R_L = 10 - 8 = 2\,\mathrm{V} \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Basic Transistor Circuit Configurations and Transistor Characteristics – “… If we simply find the ratio of IC and IB we get what is called βdc of the transistor. Hence,
\( \displaystyle \beta_{dc} = \frac{I_C}{I_B} \qquad (14.11) \)”
Transistor as a Device – “… Applying Kirchhoff’s voltage rule to the input and output sides of this circuit, we get
\( \displaystyle V_{BB} = I_B R_B + V_{BE} \qquad (14.12) \)
and
\( \displaystyle V_{CE} = V_{CC} - I_C R_C \qquad (14.13) \)”