Electrical Engineering - Series Circuits - Discussion

Ammeter is always connected in series with load. So, the answer is not given.

If we think the given question assume circuit is in series combination, between the second resistor end point and third resistor first point we can put the + positive problem and other source side direct give to negative problem.

So, ammeter shown deflection current on whole circuit, we can get total current of series circuit : I = I1+I2+I3.

So the answer is perfect is the ammeter - (negative) problem directly connected with fourth resistor and second resistor end point put + problem.

"To measure the current out of the second resistor in a circuit consisting of four resistors".

This statement tells us that all are in series because second in four means they are in one line.

It can be connected any where in the circuit because same current will be in the all resistance because of series combination.

Hence can be connected any where.

How can it be measured directly by keeping anywhere it's not mentioned as series in the question, if series it was right but in parallel?

The question is not clearly telling whether it is a series circuit or something else. So, the answers are not suitable.

In question it is not given that resistor are connected in series so option D is wrong please give clear answer.

I agree it is not mentioned that they are in series. But, we are in questions of series section, friends.

I disagree with @Kiran, because the question did not spelt out if all the resistors have the same value.

I think option "D" is not actually right, clearly there is no answer contained in the options.

Option D is correct, Because in series circuit current remain same at any point of circuit.