Answer:
a) Vout= 5V
b) Vout= 5V
c) Vbase= 0.6V
d) Vbase= 0.6V
Explanation:
Consider the circuit shown in attachment
a) When Vin is 0V, the base circuit is not turned, so
Ib=0 and Ic=∞ as transistor is not turned on so
Vout =5V
b) When Vin= 5 V,
Ib= (Vin-Vb)/Rb
Ib=(5-0.6)/1000= 0.0044A
Ic= 0.0044×10=0.044A
Vout= 5- 0.044×1000= not real value
Vout= Vce= 5V
c) voltage drop across Vbase= 0.6V
d) Vbase= 0.6V
In all the above cases, the transistor will not be turned on biasing base voltage and resistor values are very high compared to VCC which is 5V in the given circuit
Answer:
no not really
Explanation:
From your friendly neighborhood cereal killer,
Sir. BLOODPR1NCE
Answer:
b is the answer
Explanation:
after things are used don't they get replaced
Answer:
The voltage needed to accelerate the electron beam is 2.46 x 10^16 Volts
Explanation:
The rate of electron flow is given as:
q = 1015 electrons per second
The total current is given by:
Total Current = (Rate of electron flow)(Charge on one electron)
Total Current = I = (1015 electrons/s)(1.6 x 10^-19 C/electron)
I = 1.624 x 10^-16 A
Now, we know that electric power is given as:
Electric Power = Current x Voltage
P = IV
V = P/I
V = 4 W/1.624 X 10^-16 A
<u>V = 2.46 x 10^16 Volts</u>
You need to explain it more simple as everyone is clueless
