Find the resistance that must be placed in series with a 25.0Ω25.0 Ω galvanometer having a 50.0µA50.0 μA sensitivity (the same as the one discussed in the text) to allow it to be used as a voltmeter with a 0.100-V full-scale reading

1975 ohm

Explanation:

The maximum current that can pass through the galvanometer is 50.0µA

p.d across the galvanometer at a full scale deflection is given by

V = IR

V = (50.0×10^-6)*25

V1 = 1/800 or 0.00125volts

Since the p.d across the multiplier and galvanometer at full scale deflection is 0.100volts,the p.d across the multiplier alone must be given by

V2 = 0.100-0.00125

= 0.09875volts

Current passing through the multiplier is (50.0×10^-6)A

Hence if R is the resistance of the multiplier, we have that

V2 = (50.0×10^-6)×R

0.09875volts = (50.0×10^-6)×R

Divide both sides by 0.09875volts

R = 0.09875/(50.0×10^-6)

R =( 0.09875) ÷ 1/20000

R = 0.09875×20000

R = 1975 ohms

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000