Solved Numerical Problem No. 18.3 ( Unit-18: Electronics); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

In circuit (Fig.P.18.3). there is negligible potential drop between B and E. if. P is 100. Calculate 
(i) base current 
(ii) collector's current 
(iv) potential drop across `\R_C` 
(iv) `\V_{CE}`

(Ans: 11.25 μA. 1.125 mA, 1.125 V, 7.875 V) 

Given:

Voltage at Collector = `\V_{CC}` =  9 V

Voltage at Base Current = `\V_{BC}` =  0 V

Load resistance = `\R_C` = 1 kΩ = 1ｘ10³ Ω

Base resistance = `\R_B` = 800 kΩ = 800ｘ10³ Ω

Current gain = β = 100

To Find:

1. Base Current = `\I_B` = ?

2. Collector Current = `\I_C` = ?

3. Potential Drop at `\R_C` = `\V_{R_C}` = ?

4. Potential at emitter = `\V_{C_E}` = ?

Solution:

1. Base Current = `\I_B` = ?

`\V_{C_C}` = `\I_B` `\R_B`  +  `\V_{B_E}`

or 

`\I_B` = `\frac {V_{C_C} - V_{B_C} }{R_B}` 

By putting the corresponding value

`\I_B` = `\frac {9 V - 0V} }{800ｘ10⁻³ Ω}` 

`\I_B` = 0.01125ｘ10³ A

or

`\I_B` = 11.25ｘ10⁻⁶ A

or

`\I_B` = 11.25 μA ------------------Ans.1

2. Collector Current = `\I_C` = ?

Using the Current gain β  formula

β = `\frac {I_C}{I_B}` 

or

`\I_C` = β ｘ `\I_B`

putting values

`\I_C` = 100 ｘ 11.25ｘ10⁻⁶ A

`\I_C` = 11.25 ｘ10⁻⁴ A

or

`\I_C` = 1.125 ｘ10⁻³  A

or

`\I_C` = 1.125 mA --------------Ans. (2)

3. Potential Drop at `\R_C` = `\V_{Rc}` = ?

Using Ohm law 

`\V_{R_C}` = `\I_C``\R_C`

 

by putting values

`\V_{R_C}` = 1.125 ｘ10⁻³  A  ｘ1ｘ10³ Ω

`\V_{R_C}` = 1.125 ｘ10⁻³  A  ｘ1ｘ10³ Ω

`\V_{R_C}` = 1.125 V -------------------Ans.3

4. Potential at emitter = `\V_{C_E}` = ?

From the figure we have

`\V_{C_C}` = `\V_{R_C}`   +  `\V_{C_E}`

or

`\V_{C_E}` = `\V_{C_C}` - `\V_{R_C}`

By putting values

`\V_{C_E}` = 9 V - 1.125 V

`\V_{C_E}` = 7.875 V------------Ans.4

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