Question

When an ammeter of negligible internal resistance is inserted in series with
circuit it reads 1A. When the voltmeter of very large resistance is
connected across X it reads 1V. When the point A and B are shorted by a
conducting wire, the voltmeter measures 10 V across the battery.
The internal resistance of the battery is equal to

• A. zero
• B. 0.5 $\Omega$
• C. 0.2 $\Omega$
• D. 0.1 $\Omega$

Answer 1

The correct option is C 0.2 $\Omega$

$i=\frac{12}{r+R_x+R_y},r+R_x+R_y=12⟶\left(1\right)i{R}_x=1V\frac{R_x\left(12\right)}{r+R_x+R_y}=111R_x=r+R_y⟶\left(2\right)$

Now if $y$ is shorted,

$i=\frac{12}{R_x+r}$

Voltage across $x=i{R}_x$

$\frac{12R_x}{R_x+r}=10$

$2R_x=10r,R_x=5r$

Putting in $\left(1\right)R_y=12-6r$ now putting in $\left(2\right)$

$11\times 5r=r+12-6r$

$55r+5r=12$

$r=\frac{1}{5}⟹0.2\Omega$