Two identical galvanometers are converted into an ammeter and a milliammeter. For the same current,the value of the shunt of the ammeter as compared to that of the milliammeter is

The resistance of a galvanometer is $25\,\Omega$ and it requires $50\,\mu A$ for full-scale deflection. The value of the shunt resistance required to convert it into an ammeter of $5\,A$ range is:

The resistance of a galvanometer is $50\,\Omega$ and the current required to give full-scale deflection is $100\,\mu A$. In order to convert it into an ammeter for reading up to $10\,A$,it is necessary to put a resistance of:

$A$ milliammeter of range $10\, mA$ and resistance $9\, \Omega$ is joined in a circuit as shown. The meter gives full-scale deflection for current $I$ when $A$ and $B$ are used as its terminals,i.e.,current enters at $A$ and leaves at $B$ ($C$ is left isolated). The value of $I$ is

$A$ student uses the resistance of a known resistor $(1 \,\Omega)$ to calibrate a voltmeter and an ammeter using the circuits shown below. The student measures the ratio of the voltage to current to be $1 \times 10^3 \,\Omega$ in circuit $(a)$ and $0.999 \,\Omega$ in circuit $(b)$. From these measurements,the resistance (in $\Omega$) of the voltmeter and ammeter are found to be close to

An ammeter with internal resistance $90\,\Omega$ reads $1.85\,A$ when connected in a circuit containing a battery and two resistors $700\,\Omega$ and $410\,\Omega$ in series. The actual current in the circuit (without the ammeter) will be: