Question

In the diagram shown below, the cell and the ammeter both have negligible resistance. The resistors are identical. With the switch 'K' open, the ammeter reads 0.6A . What will be the ammeter reading when the switch is closed?

Answer 1

In first Situation,
Two parallel Resistance are there.
Third one will not be considered as key is open.

So Equivalent Resistance is R/2.
Using ohm's law,

V = I Req
V = 0.6 * R/2........(1)

Now in second situation,
Now all three Resistance will be considered.
So,
Equivalent Resistance is R/3

Using ohm's law,

V = I' * R/3...........(2)

(2) = (1)

I' R/3 = 0.6 * R/2

I' = 0.6 * 3/2
I' = 0.3 * 3
I' = 0.9 A........(Answer)

Answer 2

Given that,

Cell and ammeter have negligible resistance.

Reading of ammeter when switch 'K' is closed = 0.6 A

To find,

Ammeter reading when the switch is closed = ?

Actually, the question is to find the electric current. This is an indirect question. Whenever you get any question like this which says to find the ammeter reading, you have to find the electric current.

How to understand?

There is a simple logic behind this. What does a ammeter calculates? It calculates electric current. So here, we've to find electric current which is denoted by I.

Assumption:

Let the resistance of each resistor be R.

Case I:

When switch K is open only two resistors will be parallel because the electric current will not flow through the third resistor.

As the two resistors are in parallel, there equivalent resistance will be,

$\implies \frac{1}{R_1} = \frac{1}{R_1}+ \frac{1}{R_2}$

$\implies \frac{1}{R_1} = \frac{1}{R} + \frac{1}{R}$

$\implies \frac{1}{R_1} = \frac{2}{R}$

$\boxed{\bold{\mathsf{R_1 = \frac{R}{2}}}}$

Now, we'll find the value of Potential difference.

Assumption:

Let the potential difference be V volt.

The electric current = 0.6 A (as given)

By using,

V = $\bold{IR_1}$

$\implies V = 0.6 \times \frac{R}{2}$

$\implies V = \cancel{0.6} \times \frac{R}{\cancel{2}}$

$\boxed{\bold{\mathsf{V = 0.3R\:Volt}}}$

Case II:

When the switch is closed, all the three resistors will be in parallel as all the three resistors receives the electric current.

$\therefore \implies \frac{1}{R_2} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

$\implies \frac{1}{R_2} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R}$

$\implies \frac{1}{R_2} = \frac{1+1+1}{R}$

$\implies \frac{1}{R_2} = \frac{3}{R}$

$</p><p>\boxed{\bold{\mathsf{\implies R_2 = \frac{R}{3}}}}$

Now, by using Ohm's law:

V = $\bold{IR_2}$

$\implies 0.3R = I \times \frac{R}{3}$

$\implies I = \dfrac{0.3R}{\frac{R}{3}}$

$\implies I = 0.3R \times \frac{3}{R}$

$\implies I = 0.3\cancel{R} \times \frac{3}{\cancel{R}}$

$\implies I = 0.3 \times 3$

$\boxed{\bold{\mathsf{\implies I = 0.9\:A}}}$

Therefore, the ammeter's reading will be 0.9 A.

$\boxed{\bold{\pink{\mathsf{Note:}}}}$

Always remember that, in a parallel circuit, potential difference will be same and current will be different.

In a series circuit the potential difference will be different and current will be same.