Question

the resistance of an ammeter of range 5 A is 1.8ohm . a shunt of 0.2 ohm is connected in parallel to it. when it's indicater shows a current of 2 A then what will be the effective current?

Answer 1

Answer: The effective current is 20 A.

Explanation:

Given that,

Current I =  2 A

Resistance R = 1.8 ohm

Using ohm's law

The voltage is

$V= I R$

$V = 2A\times1.8\Omega$

$v = 3.6 volt$

The voltage across shunt of  0.2 ohm will be 3.6 volt because they are connected in parallel.

Therefore, the current is

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

$I' = \dfrac{3.6 volt}{0.2\Omega}$

$I' = 18 A$

Now, the effective current about the circuit

$I_E = I + I'$

$I_{E} = 20 A$

Hence, the effective current is 20 A.

Answer 2

"According to the given data, the equation used for the current and shunt with resistance is as – I’ . R  = ( I – I’  ) .

S  where I’ is the current in the ammeter and R  is the resistance of ammeter, with the I to be the effective current and S to be the shunt.

Thereby, $2\times 1.8\quad =\quad (I-2)\times 0.2$

$\Rightarrow 3.6 = 0.2 I - 0.4$

$\Rightarrow 0.2 I = 4$

$\Rightarrow I = 20 A$"