AI 11. In an electrical circuit two resistors of 2 Q and A 22, respectively are connected in series to a 5 V
battery. The heat dissipated by the 4 Q2 resistor in 5 s will be :
(a) 57
(b) 101
(c) 201.
(d) 30 J.
Answers
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0
Answer:
20 J
Explanation:
Energy dissipated = PtorVI×tor =I
2
Rt Joules
Consider Ohm's law V=IR.
In this case the resistors are connected in series and hence, the total resistance is 4+2=6 ohms. In a series circuit, the current remains the same throughout the circuit and so we use the I
2
Rt Joules formula.
The total current in the circuit is calculated as I=V/R=6/6=1ampere.
Therefore, the power dissipated P across the 4 ohm resistor for 5 s
= 1
2
×4×5=20Joules
Hence, The heat dissipated by the 4 resistor in 5 s will be 20 J.
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