Question

an electric meter which is coonected to a 220v supply line has two resitance coil a and b of 24v resistance each these coil can be used sepretaly in series or parallel . calculate the current drawn. a. only 1coila is used
b.coil a b are used in series
c. a b are used in parallel

Answer 1

a) When 1 coil is used,

Potential Difference or Voltage(V)=220 V
Resistance(R)=24Ω,  Current(I)=?

By using the formula,I=V/R
                                  I=220/24=9.17 A (Ans.)

b) When coil A and B are used,
    R(Total equivalent resistance)= 24Ω+24Ω= 48Ω
    Potential Difference or Voltage(V)=220 V  , Current(I)=?
  
      By using the formula,I=V/R
                                         I=220/48=4.08 A (Ans.)

c) When A and B are used in parallel,
     R(Total effective resistance)= (24×24)/(24+24)
                                                  =12 Ω
Potential Difference or Voltage(V)=220 V  , Current(I)=?

By using the formula,I=V/R
                                  I=220/12= 18.33 A (Ans.)