Question

2 ohm resistor and 6 ohm resistor are connected in series with a 4 V battery. Calculate amount of charge passing through the battery in 10 seconds
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Answer 1

Explanation:

 V = 4V, R1 = 6 ohm, R2 = 8 ohm (in series) (a) Combined resistance, R = R1 + R2 = 6+2 = 8ohm. (b) Current flowing, I = V/R = 4/8=0.5amp.

Answer 2

Correct option is

Correct option isC

Correct option isC20 J

Correct option isC20 JEnergy dissipated = PtorVI×tor =I

Correct option isC20 JEnergy dissipated = PtorVI×tor =I 2

Correct option isC20 JEnergy dissipated = PtorVI×tor =I 2 Rt Joules

Correct option isC20 JEnergy dissipated = PtorVI×tor =I 2 Rt JoulesConsider Ohm's law V=IR.

Correct option isC20 JEnergy dissipated = PtorVI×tor =I 2 Rt JoulesConsider 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

Correct option isC20 JEnergy dissipated = PtorVI×tor =I 2 Rt JoulesConsider 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

Correct option isC20 JEnergy dissipated = PtorVI×tor =I 2 Rt JoulesConsider 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.

Correct option isC20 JEnergy dissipated = PtorVI×tor =I 2 Rt JoulesConsider 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.

Correct option isC20 JEnergy dissipated = PtorVI×tor =I 2 Rt JoulesConsider 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

Correct option isC20 JEnergy dissipated = PtorVI×tor =I 2 Rt JoulesConsider 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

Correct option isC20 JEnergy dissipated = PtorVI×tor =I 2 Rt JoulesConsider 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

Correct option isC20 JEnergy dissipated = PtorVI×tor =I 2 Rt JoulesConsider 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

Correct option isC20 JEnergy dissipated = PtorVI×tor =I 2 Rt JoulesConsider 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=20JoulesHence, The heat dissipated by the 4 resistor in 5 s will be 20 J.

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