Electronics and Communication Engineering - Electronic Devices and Circuits - Discussion
Discussion Forum : Electronic Devices and Circuits - Section 14 (Q.No. 27)
27.
The diode and the moving coil milliammeter of figure are assumed to be ideal. The meter reading is
0.1 mA
Discussion:
5 comments Page 1 of 1.
Shreya said:
4 years ago
Since both, the diodes are in the forward bias condition so they act as a closed switch.
For a half wave rectifier,
Vdc = Vm/π.
Vm = 1V.
The resistors are in parallel so,
Req. = 200k / (20k + 10k) ohm.
By ohm's law, V = iR.
i = 1/π[3/ 20] mA.
Current through milli ammeter.
I = i*(2/3).
= 1/π(10) mA.
I = 0.1/πmA.
For a half wave rectifier,
Vdc = Vm/π.
Vm = 1V.
The resistors are in parallel so,
Req. = 200k / (20k + 10k) ohm.
By ohm's law, V = iR.
i = 1/π[3/ 20] mA.
Current through milli ammeter.
I = i*(2/3).
= 1/π(10) mA.
I = 0.1/πmA.
Paul said:
6 years ago
Reuben used 1V in CDT which is incorrect.
But It is correct in using Half wave rectifier formula.
Using Ohm's law:
I=(1V/π)/10K.
= 0.1mA.
But It is correct in using Half wave rectifier formula.
Using Ohm's law:
I=(1V/π)/10K.
= 0.1mA.
Reuben said:
7 years ago
Since half wave rectified:
Vm /π
V = Vmsin(wt).
Vm = 1.
by CDT:
(1/π) [10k / (20k + 10k)] = 0.1061/π mA or 0.1/π mA.
Vm /π
V = Vmsin(wt).
Vm = 1.
by CDT:
(1/π) [10k / (20k + 10k)] = 0.1061/π mA or 0.1/π mA.
Mukund said:
8 years ago
How? can anybody explain it.
Deepan said:
8 years ago
Explain it.
(1)
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