The coil of a galvanometer has a resistance of 18.2 Ω, and its meter deflects full scale when a current of 6.78 mA passes through it. To make the galvanometer into an ammeter, a 26.9-mΩ shunt resistor is added to it. What is the maximum current that this ammeter can read?

Explanation:

Ammeter is used in series in a circuit .It has two resistances in parallel . They are 18.2 ohm  and .0269 ohm . For full scale deflection ,  current of .00678 A passes in 18.2 ohm .

potential over it

= 118.2 x .00678

= .801396 V

Same volt will act on .0269 ohm shunt which is in parallel

current in it

.801396 / .0269 A

= 29.79 A

Total current

29.79 + .00678

= 29.79678

=

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

current flows in a wire