Question

four identical resistors each of resistance 2 ohm are
connected to battery of 20 V and internal resistance 2 ohm as given below. The reading of ideal ammeter A1. and A2, are
(1) 8A,8A
(3) 10 A, 10
(2) 5A, 5A
(4) 4A, 4 A​

Answer 1

Given that,

Four identical resistors each of resistance 2 ohm.

Voltage = 20 V

Internal resistance = 2 ohm

According to given figure,

Four resistance are connected in parallel and internal resistance connected in series.

We need to calculate the equivalent resistance

Using formula of parallel

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}}$

Put the value into the formula

$\dfrac{1}{R_{eq}}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}$

$\dfrac{1}{R_{eq}}=\dfrac{4}{2}$

$R_{eq}=\dfrac{1}{2}\ \Omega$

We need to calculate the current

Using ohm's law

$V = I(R+r)$

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

Put the value into the formula

$I=\dfrac{20}{\dfrac{1}{2}+2}$

$I=8 A$

The ideal ammeter A₁ and A₂ are connected in series.

So the current will be same .

Hence, The current in A₁ and A₂ is 8 A and 8 A.

(1) option is correct

Answer 2

Answer:

I think its 10 A ,10A

Explanation:

Here ammeter's resistence isn't given so we can assume it to be an ideal ammeter of negligible resistance.

Hence here ammeter will short and just act as connecting wire.So equivalent resistance is just due to 2 ohm internal resistance.

I=20/2 = 10A.