A light bulb with a resistance R = 75 Ohm was connected to a current source with an internal resistance r = 5 Ohm.
February 18, 2021 | education
| A light bulb with a resistance R = 75 Ohm was connected to a current source with an internal resistance r = 5 Ohm. If the amount of heat equal to Q = 1125 J is released to the light bulb in 4 minutes, then the EMF of the source is equal to.
Power P dissipated per bulb:
P = Q / t,
where Q is the amount of heat,
t is time.
The same power P, expressed in terms of resistance R and current through the bulb I:
P = I2R.
We find the current through the light bulb I:
Q / t = I2R;
I2 = Q / tR;
I = √Q / tR.
Ohm’s law for a complete circuit:
I = E / (R + r),
where E is the EMF,
r – internal resistance.
E = I * (R + r) = √ (Q / tR) * (R + r) =
= √ (1125 J / (240 s * 75 Ohm)) * (75 Ohm + 5 Ohm) =
= √ (45 / (240 * 3)) * 80V = √1 / 16 * 80 = 20V.
Answer: 20 V.
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