Question

Two ammeters X and Y have resistances of $1.2\Omega$ and $1.5\Omega$ respectively and they give full scale deflection with 150 mA and 250 mA respectively. The ranges have been extended by connecting shunt so as to give full scale deflection with 15 A.
The ammeters along with shunts are connected in parallel and then placed in a circuit in which the total current flowing is 15 A. The current in amperes indicated in ammeter X is

• A. 10.28

Answer 1

The correct option is A 10.28

Given,
$R_{mx}=1.2\Omega ,R_{my}=1.5\Omega$

$I_{mx}=0.15A,I_{my}=0.25A$

$andI=Fullscaledeflectioncurrent=15A$
$\therefore ShuntmultiplyingfactorforammeterXis{m}_X=\frac{I}{I_{mx}}=\frac{15}{0.15}=100$

and shunt multiplier resistance,

$R_{s{h}_X}=\frac{R_{m_X}}{(m_X-1)}=\frac{1.2}{99}=0.012\Omega$


Also, shunt multiplying factor for ammeter Y is
$m_Y=\frac{I}{I_Y}=\frac{15}{0.25}=60$

and shunt multiplier resistance,
$R_{s{h}_Y}=\frac{R_{m_Y}}{(m_Y-1)}=\frac{1.5}{59}=0.025\Omega$

When these ammeters are connected in parallel as shown in the figure below:



Let current in ammeter X be $I_X$ then

$R_X=\left(\frac{R_{s{h}_X}.R_{m_X}}{R_{s{h}_X}+R_{m_X}}\right)=\left(\frac{0.012\times 1.2}{1.212}\right)$

$=0.011\Omega$

and

$R_Y=\left(\frac{R_Y}{R_X+R_Y}\right)\times 15=\frac{0.024}{0.035}\times 15=0.024\Omega$

$\therefore I_X=\left(\frac{0.024}{0.024+0.011}\right)\times 15$

$\left(\frac{R_Y}{R_X+R_Y}\right)\times 15=\frac{0.024}{0.035}\times 15=10.28A$

$\therefore CurrentinammeterX=10.28A$