Answer

Verified

20.4k+ views

**Hint:**The main supplies a potential difference which originates a current flow through the circuit. The bulbs on the other hand use the current flow through them to glow. The resistance in the bulb causes some voltage drop and eventually reduces the current flow. The ammeter in the circuit takes the reading of the current flow in the circuit.

**Formula Used:**

If a circuit has a source of potential $V$ and an effective resistance $R$ then the current flow through the circuit can be calculated as

$I = \dfrac{V}{R}$

**Complete step by step answer:**

Step 1:

You can see from the figure that the ammeter $A$ notes the reading of the current through the circuit.

You can ideally take the resistance of the ammeter to be $0$.

Hence, there will be no potential drop across the ammeter $A$ .

So, from eq (1), the reading of the ammeter can be defined as

$I = \dfrac{{{V_{main}}}}{{{R_{circuit}}}}$

where the main supplies a potential difference of ${V_{main}}$ and ${R_{circuit}}$ is the equivalent resistance of the circuit.

Step 2:

Let the main supplies a potential difference of ${V_{main}} = V$

When the switch is open, the circuit is connected to only the bulb $P$ .

Let the bulb have a resistance $R$ and hence the effective resistance in the circuit is ${R_{circuit}} = R$.

Hence, calculate the reading ${I_1}$ of the ammeter $A$ from the eq (2)

$

{I_1} = \dfrac{{{V_{main}}}}{{{R_{circuit}}}} \\

\Rightarrow {I_1} = \dfrac{V}{R} \\

$

Therefore, the reading of the ammeter is ${I_1} = \dfrac{V}{R}$.

Step 3:

When the switch is closed, the circuit is connected to the bulb $P$ and the bulb $M$ parallelly.

By the problem, $P$ and $M$ are two identical bulbs.

Hence, the bulb has a resistance $R$ .

Calculate the effective resistance in the circuit.

$

\dfrac{1}{{{R_{circuit}}}} = \dfrac{1}{R} + \dfrac{1}{R} \\

\Rightarrow \dfrac{1}{{{R_{circuit}}}} = \dfrac{2}{R} \\

\Rightarrow {R_{circuit}} = \dfrac{R}{2} \\

$

Hence, calculate the reading of the ammeter $A$ from the eq (2)

$

{I_2} = \dfrac{{{V_{main}}}}{{{R_{circuit}}}} \\

\Rightarrow {I_2} = \dfrac{V}{{\dfrac{R}{2}}} \\

\Rightarrow {I_2} = 2\dfrac{V}{R} \\

$

Therefore, the reading of the ammeter is ${I_2} = 2\dfrac{V}{R}$.

Step 4:

Hence, compare the two readings of the ammeter.

${I_2} = 2\dfrac{V}{R} = 2{I_1}$

$\therefore {I_2} = 2{I_1}$.

Final Answer:

If the ammeter $A$ is affected by connecting another identical bulb $Q$ in parallel to $P$ with the voltage in the mains, maintained at a constant value, then (C) The reading will be double the previous value.

**Note:**The identical bulbs should have the same resistance. You should neglect the internal resistance of the wires or the ammeter for the simplicity of the calculation. When the switch is open, the bulb $M$ shall not be connected to the circuit, but when the switch is closed, you should take the parallel connection as mentioned in the problem. The resistance of the bulbs can only cause a potential drop else, current flow through them remains unchanged.

Recently Updated Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

The path difference between two waves for constructive class 11 physics JEE_MAIN

What is the difference between solvation and hydra class 11 chemistry JEE_Main

IfFxdfrac1x2intlimits4xleft 4t22Ft rightdt then F4-class-12-maths-JEE_Main

Three point particles of mass 1 kg 15 kg and 25 kg class 11 physics JEE_Main

Sodium chloride is purified by passing hydrogen chloride class 11 chemistry JEE_Main

Other Pages

The phase difference between the particles vibrating class 11 physics JEE_Main

Chloroform reacts with oxygen in the presence of light class 12 chemistry JEE_Main

Give one chemical test to distinguish between the following class 12 chemistry JEE_Main

Assertion The melting point Mn is more than that of class 11 chemistry JEE_Main

The correct order increasing acidic strength is A Phenol class 12 chemistry JEE_MAIN

Calculate CFSE of the following complex FeCN64 A 04Delta class 11 chemistry JEE_Main