A galvanometer of resistance $$40\Omega$$ gives a deflection of $$5$$ divisions per $$mA$$. There are $$50$$ divisions on the scale. The maximum current that can pass through it when a shunt resistance of $$2\Omega$$ is connected is:
$$\displaystyle I_g$$=$$\dfrac{50}{5}=10mA$$;
$$R_g$$ = 40 $$\Omega$$, $$R_s$$= 2$$\Omega$$
Maximum current,
$$I=\dfrac{R_g+R_s}{R_g}$$ X $$I_g$$ = $$\dfrac{(40+2)\times10}{2}$$ = $$210mA$$
A voltmeter coil has resistance $$50.0 \Omega$$ and a resistor of $$1.15 \ k\Omega$$ is connected in series, It can read potential difference an ammeter which $$12 volts$$. If this same coil is used to construct an ammeter which can measure currents up to $$2.0 A$$, what should be the resistance of the shunt used?
An ammeter is to be constructed which can read currents up to $$2.0 A$$. If the coil has a resistance of $$25 \Omega$$ and takes $$1 mA$$ for full-scale deflection, what should be the resistance of the shunt used?
An ammeter gives full scale deflection with a current of $$1\ amp$$. It is converted into an ammeter of range $$10\ amp$$. The ratio of the resistance of ammeter to the shunt resistance used will be :
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The resistance of $$1\ A$$ ammeter is $$0.08 \Omega$$. To convert it into $$10\ A$$ ammeter, the shunt resistance required will be
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A galvanometer has a resistance of $$25 \ ohm$$ and a maximum of $$0.01\ A$$ current can be passed through it. In order to change it into an ammeter of range $$10\ A$$, the shunt resistance required is
A $$100\ ohm$$ galvanometer gives full scale deflection at $$10\ mA$$. How much shunt is required to read $$100\ mA$$