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Electric Components and Circuit Diagram

The statement...

Question

The statement that is most correct about the following circuit is:

$A_{1}, A_{2}, & A_{3}$ are readings of ammeter.

A

A_{3} < A_{2}

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B

A_{1} = A_{2} = A_{3}

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C

A_{1} < A_{2}

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D

A_{1} = A_{3}

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Solution

The correct option is B $A_{1} = A_{2} = A_{3}$
Current in components in series is constant. For the middle part of the circuit, ammeter is connected in parallel to the resistance R. Due to its extremely low resistance, all the current will pass through the ammeter and middle resistance can be ignored. Hence, all the ammeters will read the same.

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Similar questions

a_{1}, a_{2}, a_{3}, . . . a_{n}

are in H.P then

\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . a_{n}}, \frac{a_{3}}{a_{1} + a_{2} + a_{4} + . . . + a_{n}}, \frac{a_{n}}{a_{1} + a_{2} + . . . + a_{n}}

\frac{a_{1}}{a_{1}} + \frac{a_{2}}{a_{2}} + \frac{a_{3}}{a_{3}} + \dots \frac{a_{n}}{a_{n}} = \frac{1}{1} + \frac{1}{a_{1}} + \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \dots \frac{a_{n - 2}}{a_{n - 1}}

a_{1}, a_{2}, a_{3}, . . . . ., a_{n}

are in H.P. then prove that

\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{3}}{a_{2} + a_{3} + . . . a_{n}}

a_{1}, a_{2}, a_{3}, . . . . a_{n}

are in H.P, then

\frac{a_{1}}{a_{2} + a_{3} + . . . . + a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . . + a_{n}}, . . . ., \frac{a_{n}}{a_{1} + a_{2} + . . . . + a_{n - 1}}

a_{1}, a_{2}, a_{3}, . . . . . a_{n}

are in H.P., then

\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . . + a_{n}}, \frac{a_{n}}{a_{1} + a_{2} + . . . + a_{n - 1}}

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