$A$ galvanometer gives a full-scale reading of $50 \ mA$,when a potential difference across its terminals is $0.15 \ V$. It can be used as an ammeter of range $0 - 100 \ A$ by connecting a shunt resistance of: (in $\Omega$)

In the adjoining circuit diagram,the readings of the ammeter and voltmeter are $2 \, A$ and $120 \, V$,respectively. If the value of $R$ is $75 \, \Omega$,then the voltmeter resistance will be (in $\Omega$):

The sensitivity of a galvanometer is $60 \text{ division/A}$. When a shunt is used,its sensitivity becomes $10 \text{ division/A}$. If the galvanometer is of resistance $20 \ \Omega$,the value of shunt used is (in $\Omega$)

The sensitivity of a galvanometer that measures current is decreased by $\frac{1}{40}$ times by using a shunt resistance of $10 \Omega$. Then,the value of the resistance of the galvanometer is (in $\Omega$)

$A$ galvanometer coil has a resistance of $12\; \Omega$ and the meter shows full-scale deflection for a current of $3\; mA$. How will you convert the meter into a voltmeter of range $0$ to $18\; V$?

$A$ 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