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.

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Solution

a) V=IR

I=V/R

I= 220/24

I= 9.16 A

B) THEY ARE CONNECTED IN SERIES HENCE THEIR TOTAL RESISTENCE WILL BE:- 24+24= 48

I = V/R

I = 220/48

I = 4.583 A

C) THEY ARE CONNECTED IN PARALLEL HENCE THEIR TOTAL RESISTANCE WILL BE:- 1/24 + 1/24 = 1/12 = 12

V= IR

I = V/R

I = 220/12

I = 18.33 A

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a) V=IR I=V/R I= 220/24 I= 9.16 A B) THEY ARE CONNECTED IN SERIES HENCE THEIR TOTAL RESISTENCE WILL BE:- 24+24= 48 I = V/R I = 220/48 I = 4.583 A C) THEY ARE CONNECTED IN PARALLEL HENCE THEIR TOTAL RESISTANCE WILL BE:- 1/24 + 1/24 = 1/12 = 12 V= IR I = V/R I = 220/12 I = 18.33 A

a) V=IR

I=V/R

I= 220/24

I= 9.16 A

B) THEY ARE CONNECTED IN SERIES HENCE THEIR TOTAL RESISTENCE WILL BE:- 24+24= 48

I = V/R

I = 220/48

I = 4.583 A

C) THEY ARE CONNECTED IN PARALLEL HENCE THEIR TOTAL RESISTANCE WILL BE:- 1/24 + 1/24 = 1/12 = 12

V= IR

I = V/R

I = 220/12

I = 18.33 A