an ideal battery with E = 12V is connected in series to two bulbs with resistance R1 = 2ohm and R2 = 4ohm . what is the current in the circuit and the power dissipation in each bulb
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hope this will help you out...
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Anonymous:
cant we take v square by R
yes but it become complexive because you have to find V of each bulb
you can do this bulb 1 V=4
bulb 2 V=8
i hope this is correct
ya
Answered by
3
Answer:
V=I × R
12= i ×(2+4)
i =2 A
Power dissipated
a. 2R==> P=I^2×R=2×2×2= 8 watt
b.4R==> P= I^2×R =2×2×4 =16 Watt
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