Question

In the figure ammeter $\left(I\right)$ reads a current of $10mA$, while the voltmeter reads a potential difference of $3V$. What does ammeter $\left(II\right)\left(inmA\right)$ read?
The ammeters are identical, the internal resistance of the battery is negligible. (Consider all ammeters and voltmeters as non ideal)

• A. 3.33
• B. 6.67
• C. 16.67
• D. 33.33

Answer 1

The correct option is B 6.67

Voltage drop across voltmeter is $3V$ and emf is $4V$.

Hence, voltage across $A_I$ is $1V$.

Hence, resistance of ammeters is $100\Omega$.

Let the potential of the point of intersection of the $100\Omega$ resistances and $A_{II}$ be $x$.

Using KCL:

$\frac{x-4}{100}+\frac{x}{100}+\frac{x-1}{100}=0$

$3x=5$

$\Rightarrow x=\frac{5}{3}$

Hence, current in $A_{II}=\frac{x-1}{100}=\frac{2}{300}=6.67$ mA