Electronic Devices - Diode Applications - Discussion
Discussion Forum : Diode Applications - General Questions (Q.No. 32)
32.
Determine the value of the load resistor.
RL = 5 k
RL = 5.5 k
RL = 6 k
None of the above
Discussion:
7 comments Page 1 of 1.
Pavan said:
1 decade ago
RL=12v-(0.3+0.7)v/2mA=5.5Kohm
ge=0.3v & si=0.7v
ge=0.3v & si=0.7v
Ashok said:
1 decade ago
Dear, @Pawan.
I solved, this equation [RL=12v-(0.3+0.7)v/2mA],
But, I am not able to get 5.5 Kohm.
Can you explain it.
I solved, this equation [RL=12v-(0.3+0.7)v/2mA],
But, I am not able to get 5.5 Kohm.
Can you explain it.
RAJUSRIPATHI said:
1 decade ago
RL = (12-(0.3+0.7))/2, Do it you will get:
RL = (12-1)/2.
RL = 11/2.
RL = 5.5 OHM.
RL = (12-1)/2.
RL = 11/2.
RL = 5.5 OHM.
Madhu said:
1 decade ago
How ge=0.3v & si=0.7V?
Abishek.M said:
1 decade ago
si = 0.7 v, ge = 0.3 v.
RL= (12-(0.3+0.7))/2.
Rl = (12-1.0)/2.
RL = 11/2.
Rl = 5.5 ohm.
RL= (12-(0.3+0.7))/2.
Rl = (12-1.0)/2.
RL = 11/2.
Rl = 5.5 ohm.
Rajkumar said:
9 years ago
RL = (12 - 0.3 - 0.7)/2mA.
= (12 - 1.0)/2 *10^-3.
= (11 * 10^3)/2.
= 5.5 * 10^3.
= 5.5 kohm.
= (12 - 1.0)/2 *10^-3.
= (11 * 10^3)/2.
= 5.5 * 10^3.
= 5.5 kohm.
Jitendra Chauhan said:
9 years ago
Si = 0.3 VDC.
Ge = 0.7 VDC.
Network Voltage = 12 VDC.
We know that R = V/I ,
So total V = [12 - (0.3 + 0.7)] = 11.
I = 2 mA.
R= 11/2 = 5.5 ANS.
Ge = 0.7 VDC.
Network Voltage = 12 VDC.
We know that R = V/I ,
So total V = [12 - (0.3 + 0.7)] = 11.
I = 2 mA.
R= 11/2 = 5.5 ANS.
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