Q20. Two resistors of resistance 2 ohm and 4 ohm are connected in series with the battery of 6 volt with an ammeter to measure the current through 4 resistor and voltmeter to measure the voltage across 2 ohm resistor with the
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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. please mark me as brailist
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