Question

A circuit consists of a battery B of emf 10 V, three resistors X, Y, Z of

resistance 10 Ω, 10 Ω and 5 Ω respectively, ammeters A₁ and A₂, a voltmeter

V₁ and a key K. A₁ reads 1 A, A₂ reads 0.5 A and V₁ reads 5 V. Draw a

diagram to show the circuit arrangement. What will happen if connection to

X is left open?
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Answer 1

Answer:-

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Answer 2

Given:

Value of resistors are 10 Ω, 10 Ω and 5 Ω respectively.

A₁ reads 1 A, A₂ reads 0.5 A and V₁ reads 5 V.

To find:

Diagram to show the circuit arrangement. What will happen if connection to  X is left open.

Explanation:

For the initial conditions, the circuit arrangement will be as given in fig. (A).

When connection to X is left open, the circuit will be as shown in fig. (B).

Then, reading of $A_{1}$ and $A_{2}$ will be,

                      $I = \frac{V}{R_{eq}} =\frac{10}{10+5}=\frac{10}{15}=\frac{2}{3} A$

Voltmeter reads potential drop across the 5Ω resistor.

Therefore, voltmeter reading will be,

                     $V_{1}=\frac{10}{10+5}\times5=\frac{10}{3} V$