Electrical Engineering - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 13)
13.
When a load resistance is removed from the output of a voltage divider circuit, the current drawn from the source
Discussion:
20 comments Page 2 of 2.
SanKau PM said:
10 years ago
Load resistance is removed from the output of a voltage divider circuit. How the less current will flow in open circuit. Me too think than current will cut off?
Santy said:
1 decade ago
In voltage divider circuit the load is connected across a resistor.
Arun said:
1 decade ago
Electrical Source in always infinite. Load cannot alter the source current. Current remains the same. But only the drop matters.
Mohan said:
1 decade ago
For example use a variable resistor and decrease range of resistor value. Now the source current would decrease.
Ankit Tiwari said:
1 decade ago
Why option D is not correct please tell me with explanation?
Niharika said:
1 decade ago
Shouldn't it remain same as the other resistor will be present Because as per the v/g divider circuit I suppose there must be more than a resistor?
NIVI said:
1 decade ago
If circuit is open then current is not flowing in the ckt, then why not its answer is D?
Biswajit said:
1 decade ago
If the load resistance will be removed then the circuit is behave like a open circuit. Which current will flow, so the drawing current will decreased.
Preethi said:
1 decade ago
If the resistance is removed then it is open circuit.
O.C current will be low, voltage will be high.
S.C current will be high, voltage will be high.
O.C current will be low, voltage will be high.
S.C current will be high, voltage will be high.
Vimal said:
1 decade ago
The Load resistance is removed.
If there is any load, it will draw some current.
If there is no load (ie, load resis), it will not draw any current from source.
So the drawing current will be decreased.
If there is any load, it will draw some current.
If there is no load (ie, load resis), it will not draw any current from source.
So the drawing current will be decreased.
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