Question:
In a transistor connected in CE configuration, the emitter current increases from 10 mA to 20 mA when the base current increases from 0.12 mA to 0.24 mA. Then the current gain is
In a transistor connected in CE configuration, the emitter current increases from 10 mA to 20 mA when the base current increases from 0.12 mA to 0.24 mA. Then the current gain is
Show Hint
Use $\beta = \dfrac{\Delta I_C}{\Delta I_B}$ for current gain in CE configuration.
Updated On: May 12, 2025
- 41.3
- 82.3
- 80.2
- 96.4
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
In CE (common emitter) configuration: $I_E = I_B + I_C \Rightarrow I_C = I_E - I_B$
Initial: $I_E = 10$ mA, $I_B = 0.12$ mA ⇒ $I_C = 9.88$ mA
Final: $I_E = 20$ mA, $I_B = 0.24$ mA ⇒ $I_C = 19.76$ mA
Current gain $\beta = \dfrac{\Delta I_C}{\Delta I_B} = \dfrac{19.76 - 9.88}{0.24 - 0.12} = \dfrac{9.88}{0.12} \approx 82.3$
Initial: $I_E = 10$ mA, $I_B = 0.12$ mA ⇒ $I_C = 9.88$ mA
Final: $I_E = 20$ mA, $I_B = 0.24$ mA ⇒ $I_C = 19.76$ mA
Current gain $\beta = \dfrac{\Delta I_C}{\Delta I_B} = \dfrac{19.76 - 9.88}{0.24 - 0.12} = \dfrac{9.88}{0.12} \approx 82.3$
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Semiconductor electronics: materials, devices and simple circuits Questions
- The p-n junction which generates an emf when the solar radiation incidents on it is
In the circuit, $ E_1 = E_2 = E_3 = 2 \, \text{V} $ and $ R_1 = R_2 = 4 \, \Omega $. Then the current flowing through $ E_2 $ is:
- n-type semiconductor is formed when germanium is doped with:
- A reverse biased zener diode when operated in the breakdown region works as
- The current amplification factor and the collector current in a transistor circuit are 100 and 2.2 mA respectively. Then the base current is
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions