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Question

An electric heater that is connected to a 220V 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 (a) Only one coil A is used and (b) coils A and B are used in series.


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Solution

Step 1: Given data

  1. The supplied voltage, V=220volt..
  2. The resistance of two coils is, 24Ω.

Step 2: Formulae

  1. We know, that current is the number of charges passing through a conductor in unit time.
  2. The current flowing through a circuit of resistance is defined by the form, I=VR, where, V is the voltage across the conductor.
  3. We know when two resistance R1 and R2 are connected in series, then the equivalent resistance of the circuit is R1+R2. If the two resistances are connected in parallel combination, then the equivalent resistance is R1R2R1+R2.

Step 3: Current through the heater when only A resistance is connected

As we know, I=VR

So,

I=VR=22024=9.61orI=9.61A.

Step 4: Diagram

An electric heater which is connected to a 220 V supply class 11 physics  CBSE

Step 5: Current through the heater when A and B both resistances are connected in series

Now, the equivalence resistance in this case is, R=R1+R2

So,

R=24+24=48orR=48Ω

Therefore, I=VR

So,

I=VR=22048=4.58orI=4.58A.

Therefore, (a) the current through the coil, when only A resistance is connected is 9.61A and (b) the current through the coil, when A and B both resistance are connected are series is 4.58A.


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