Question

please show your work.

1.) An electric blanket is connected to a 240-V outlet and consumes 111 W of power. What is the resistance of the heater wire in the blanket?
_______?

2.)A galvanometer with a coil resistance of 9.07 ? is used with a shunt resistor to make a nondigital ammeter that has an equivalent resistance of 0.41 ?. The current in the shunt resistor is 2.75 mA when the galvanometer reads full scale. Find the full-scale current of the galvanometer.
_____________A

Answer 1

1)Given that,

V = 240Volts ; P = 111 Watts

We need to find the resistance, Let it be R.

We know that,

Power = V² / R => R = V² / P = (240)² / 111 = 518.92 Ohms

Hence, Resistance of the heater wire in the blanket = R = 518.92 Ohms.

2)Given that,

R = 9.07 Ohm ; r(eq)= 0.41 Ohm, i(shunt)= 2.75 mA = 0.00275 A

Let the full scale galvanometer current be I.

shunt resistor value will be given by

r(shunt) = 1/(1/0.41 - 1/9.07) = 0.42954 Ohm

we have both r(shunt) and I(shunt) soV(shunt) will be

V(shunt) = I(shunt) x r(shunt) = 0.00275 x 0.42954 = 0.001181 Volts

Now the full scale current will be

I = V/R = 0.001181/ 9.07 = 0.00013021 Ampere = 0.130 mA = 130 micro Amperes

Hence, full scale current of the galvanometer = I = 130 A