Question

an electric appliance connected to 220v line has two resistance coils p and q each of 44ohm resistance which may be used to separate li in series or in parallel what are the current in three cases​

Answer 1

Explanation:

when the resistance p and q is connected in series the current flowing through them is 2.5A using ohms law

l=V/R

when resistances p and q are connected in parallel the current flowing them is 0.113636

Answer 2

Explanation:

An electric heater which is connected to a 220 V supply line has two resistance coils. A and B of 24Ω resistance each. these coils can be used separately (one at a time), in series or in parallel. calculate the current drawn when:

Only one coil A is used

coils A and B are used in series

When Series

R = R1 + R2

R = 24 + 24

R = 48 ohm

V = R × I ( Ohm's law )

I = V / R

I = 220 / 48 = 4.583333333 Amp

➡When parallel

1 / R = 1 / 24 + 1 / 24

1 / R = 2 / 24

1 / R = 1 / 12

R = 12 ohm

I = V / R

I = 220 / 12

I = 18.33333333 Amp