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 RC 
(iv) VCE

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



Given:

Voltage at Collector = Vℂ =  9 V

Voltage at Base Current = VBC =  0 V

Load resistance = RC = 1 kΩ = 1ｘ10³ Ω

Base resistance = RB = 800 kΩ = 800ｘ10³ Ω

Current gain = β = 100



To Find:

1. Base Current = IB = ?

2. Collector Current = IC = ?

3. Potential Drop at RC = VRC = ?

4. Potential at emitter = VCE = ?



Solution:

1. Base Current = IB = ?


VCC = IB RB  +  VBE

or 

IB = VCC-VBCRB 

By putting the corresponding value

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

IB = 0.01125ｘ10³ A

or

IB = 11.25ｘ10⁻⁶ A

or

IB = 11.25 μA ------------------Ans.1




2. Collector Current = IC = ?


Using the Current gain β  formula


β = ICIB 

or

IC = β ｘ IB

putting values

IC = 100 ｘ 11.25ｘ10⁻⁶ A

IC = 11.25 ｘ10⁻⁴ A

or

IC = 1.125 ｘ10⁻³  A

or

IC = 1.125 mA --------------Ans. (2)



3. Potential Drop at RC = VRc = ?

Using Ohm law 

VRC = ICRC
 
by putting values

VRC = 1.125 ｘ10⁻³  A  ｘ1ｘ10³ Ω

VRC = 1.125 ｘ10⁻³  A  ｘ1ｘ10³ Ω

VRC = 1.125 V -------------------Ans.3



4. Potential at emitter = VCE = ?

From the figure we have

VCC = VRC   +  VCE

or

VCE = VCC - VRC

By putting values


VCE = 9 V - 1.125 V

VCE = 7.875 V------------Ans.4



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