In the figure, the ammeter $\left(I\right)$ reads a current of $10 \mathrm{m}\mathrm{A}$, while the voltmeter reads a potential difference of $3 \mathrm{V}$. The ammeters are identical, and the internal resistance of the battery is negligible (consider all ammeters and voltmeters as non-ideal).
The resistance of the ammeter is, $m\times {10}^2 \mathrm{\Omega }$. What is the value of $m$?

In the circuit shown in figure, the reading of ideal ammeter connected as shown is same with both switches open as with both closed. Then, find the resistance $R$ (in $\mathrm{\Omega }$).

One of the circuits for the measurement of resistance by potentiometer is shown. The galvanometer is connected at point $A$ and zero deflection is observed at length $PJ=30 \mathrm{c}\mathrm{m}$. In second case, the secondary cell is changed. Take ${\epsilon }_{\mathrm{s}}=10 \mathrm{V}$ and $r=1 \mathrm{\Omega }$ in $1^{\mathrm{s}\mathrm{t}}$ reading and ${\epsilon }_{\text{s}}=5 \mathrm{V}$ and $r=2 \mathrm{\Omega }$ in $2^{\mathrm{n}\mathrm{d}}$ reading. In the second case, the zero deflection is observed at length $PJ=10 \mathrm{c}\mathrm{m}$. What is the resistance $R$ (in ohm)?

For the galvanometer of the potentiometer shown in the figure, the balance point is obtained at a distance $75 \mathrm{c}\mathrm{m}$ from $A$ when the key is open.

The second balance point is obtained at $60 \mathrm{c}\mathrm{m}$ from $A$ when the key is closed. Find the internal resistance (in $\mathrm{\Omega }$) of the battery ${\epsilon }_1$.