Question

In the circuit diagram given below, an ammeter reads 0.45 A. If the resistance Pis removed from the
circuit. The magnitude of current shown by ammeter is:

Answer 1

Given:

In the circuit diagram given below, an ammeter reads 0.45 A.

To find:

Magnitude of current when resistance P is removed?

Calculation:

Let the battery produce potential difference of V volt.

Now, net current through circuit is :

$i = \dfrac{V}{P + Q} = 0.45 \: amp$

Now, when resistance P is removed, net resistance becomes Q :

$i_{new}= \dfrac{V}{Q}$

• Since P > 0 ohm, we can say:

$\implies i_{new} > i$

$\implies i_{new} > 0.45 \: amp$

So, magnitude of current as shown by ammeter will be greater than 0.45 Amp.

Answer 2

Answer:

Given:

In the circuit diagram given below, an ammeter reads 0.45 A.

To find:

Magnitude of current when resistance P is removed?

Calculation:

Let the battery produce potential difference of V volt.

Now, net current through circuit is :

i = \dfrac{V}{P + Q} = 0.45 \: ampi=

P+Q

V

=0.45amp

Now, when resistance P is removed, net resistance becomes Q :

i_{new}= \dfrac{V}{Q}i

new

=

Q

V

Since P > 0 ohm, we can say:

\implies i_{new} > i⟹i

new

>i

\implies i_{new} > 0.45 \: amp⟹i

new

>0.45amp

So, magnitude of current as shown by ammeter will be greater than 0.45 Amp