a room heater is connected in serise with a 100w bulb.if the
100w bulb is replaced by a 40w bulb then the output of the
heater is increase/decrease/remain same?
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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
OUTPUT DECREASES
| Is This Answer Correct ? | 54 Yes | 7 No |
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