Answer Posted / vivek anand

Let us understand by taking some example.

let say heater is of 500w and 250v

Bulb 100w and 250v

now, calculate the R

P = V^2/R

SO

R = V^2/P

R1 (heater) = 250 * 250 /500 = 125 ohm

R2 (bulb)= 250 * 250 /100 = 625 ohm

Now, as the question says, they are connected in series, lets find out the current drawn by the circuit

I = V/R

LET THE CIRCUIT CONNECTED WITH 250V SUPPLY

I = 250/(125+625) (R ARE IN SERIES)

I = 0.33A

THIS IS THE AMOUNT OF CURRENT DRAWN BY CIRCUIT IN CONFIGURATION 1.

power of heater = I^2*R*T (TAKE T = 1 hr)

P = 0.33*0.33*125*60*60 = 495 ~ 500w

Now, replace 100w bulb with 40w bulb

R2 (bulb)= 250 * 250 /40 = 1562.5 ohm

New Resistance of circuit will be

R'= 125 + 1562.5 = 1687.5

**NOTE "THE CURRENT IN SERIES IS ALWAYS SAME" and "THE CURRENT DRAWN BY THE CIRCUIT DEPENDS ON LOAD (RESISTANCE)"

In this case we change the load form 750 ohm to 1687.5 ohm, so definitely the current drawn will change.

Now, I' = 250/1687.5 = 0.148 A

now this current will same in the circuit for both bulb and heater bcoz they are in series.

NEW POWER OF HEATER

P = 0.15*0.15*125*60*60 = 101.25

so

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