Ammeter

In the circuit shown, $R=10 \mathrm{\Omega }$ and $V=100 \mathrm{V}$. With the switch $S$ open the reading of the ammeter is one-third its reading when $S$ is closed. Calculate the resistance of the ammeter (in $\mathrm{\Omega }$).

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

In the circuit shown, battery, ammeter and voltmeter are ideal and the switch $S$ is initially closed as shown. When switch $S$ is opened,

A circuit contains an ideal battery, three resistors and two ideal ammeters. The ammeters read $0.2 \mathrm{A}$ and $0.3 \mathrm{A}$. After two of the resistors are switched, the readings of the ammeters did not change. Find the battery current.

The galvanometer shown in the figure has resistance $10 \mathrm{\Omega }.$ It is shunted by a series combination of a resistance $S=1 \mathrm{\Omega }$ and an ideal cell of emf $2 \mathrm{V}.$ A current $2 \mathrm{A}$ passes as shown:

Three ammeters $A, B$ and $C$ of resistances $R_A, R_B$ and $R_C,$ respectively, are joined as shown in the figure.

When some potential difference is applied across the terminals $T_1$ and $T_2,$ their readings are $I_A, I_B$ and $I_C,$ respectively, then

Two ideal batteries and two ammeters are arranged as shown in the figure.

In the above figure, if the polarity of $E_2$ is reversed, then

Two ideal batteries and two ammeters are arranged as shown in the figure.

A non-ideal ammeter and voltmeter are connected in series with an ideal cell of emf $E=22 \mathrm{V}$ as shown in the figure. Reading of the voltmeter is $V_0.$ When resistance is added in parallel to the voltmeter, its reading becomes $V_0/10$ and the reading of the ammeter becomes $10$ times the earlier value. Find the value of $V_0$ (in $\mathrm{volts}$). 

Find the reading of the ideal ammeter connected in the given circuit. Assume that the cells have negligible internal resistance.

In circuit shown in figure, terminal $X$ is brought from point $A$ to $B$. Then reading of ammeter

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

In the circuit shown, the reading of the ammeter is doubled after the switch is closed. Each resistor has the resistance $R=1 \mathrm{\Omega }$ and the ideal cell has an emf $\epsilon$ of $10 \mathrm{V}$. Then, the ammeter

In the given arrangement, the reading of ammeter is same in each case when either $K_1$ or $K_2$ is closed. The reading of the ammeter is,

Calculate the energy stored in the capacitor of capacitance $2 \mathrm{\mu }\mathrm{F}$. The voltmeter gives a reading of $15 \mathrm{V}$ and the ammeter $\mathrm{A}$ reads $15 \mathrm{mA}$.

A milli-ammeter of range $10 \mathrm{mA}$ and resistance $9 \mathrm{\Omega }$ are joined in a circuit as shown in the figure. The meter gives full-scale deflection, when the current in the main circuit is $I$, $A$ and $D$ are used as terminals. The value of $I$ is

In the given circuit (figure), when key $K$ is open, the reading of the ammeter is $I$. Now the key $K$ is closed, then the correct statement is-

In the given circuit (figure) in which case will the ammeter reading not change when $R_2$ is varied?

When a galvanometer is shunted with a $4 \mathrm{\Omega }$ resistance, the deflection is reduced to $\frac{1}{5}$. If the galvanometer is further shunted with a $2 \mathrm{\Omega }$ wire, the new deflection will be (assuming the main current remains the same),

An ammeter is obtained by shunting a $30 \mathrm{\Omega }$ galvanometer with a $30 \mathrm{\Omega }$ resistance. What additional shunt should be connected across it to double the range?