An electric heater which is connected to a 220 V supply line has two resistance coils A and B of 24 Ω resistance each. These coils can be used separately (one at a time), in series or in parallel. Calculate the current drawn when:
(a) only one coil A is used.
(b) coils A and B are used in series.
(c) coils A and B are used in parallel.
Given V = 220V
RA = RB = 24 ohm
(a) Current drawn when only coil A is used:
I = \(\frac{1}{R_A}\)
= \(\frac{220}{24}\)
= 9.16 amps
(b) Current drawn when coils A and B are used in series:
Total resistance, R = RA + RB
= 24 + 24 = 48 ohms
I = \(\frac{V}{R}\)
= \(\frac{220}{48}\)
= 4.58 amps
(c) Current drawn when coils A and B are used in parallel:
Total resistance, \(\frac{1}{R}\) = \(\frac{1}{R_A}\) + \(\frac{1}{R_B}\)
= \(\frac{1}{24}\) + \(\frac{1}{24}\)
= \(\frac{2}{24}\) = \(\frac{1}{12}\)
R = 12 ohms
I = \(\frac{V}{R}\) = \(\frac{220}{12}\)
= 18.33 amps