In an electrical circuit, two resistor of2Ω and 4Ω respectively are connected in a series of a 6V
battery. The heat dissipated by the 4Ω resistor in 5s will be
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Answered by
1
Answer:
Heat dissipated in 4Ω resistor is 20J
Explanation:
In series combination,
Net resistance=R1+R2
Net resistance=2Ω+4Ω =6Ω
V=IR
I=V/R= 6/6=1A
Heat dissipated =I^2 RT
Heat dissipated in 4Ω resistor= 1^2 x4x5=20J
Answered by
0
Answer:
Heat dissipated in 4Ω resistor is 20J
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