Question

In an electric circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6V battery. The heat dissipated by the 2 Ω resistor in 5S will be

Answer 1

Energy dissipated $\large\rm { = Pt = I^{2} Rt}$ Joules

Consider Ohm's law $\large\rm { V=IR. }$

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 $\large\rm { I²Rt }$ Joules formula.

The total current in the circuit is calculated as $\large\rm { I = V/R = 6/6 = 1 \ A}$

Therefore, the power dissipated P across the 4 ohm resistor for 5 s

$\large\rm { = 1^{2} × 4 × 5 = 20 }$ Joules

so, answer is 20 Joules

Answer 2

Answer:

Energy dissipated \large\rm { = Pt = I^{2} Rt}=Pt=I

Energy dissipated \large\rm { = Pt = I^{2} Rt}=Pt=I 2

Energy dissipated \large\rm { = Pt = I^{2} Rt}=Pt=I 2 Rt Joules

Energy dissipated \large\rm { = Pt = I^{2} Rt}=Pt=I 2 Rt JoulesConsider Ohm's law \large\rm { V=IR. }V=IR.

Energy dissipated \large\rm { = Pt = I^{2} Rt}=Pt=I 2 Rt JoulesConsider Ohm's law \large\rm { V=IR. }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 \large\rm { I²Rt }I²Rt Joules formula.

Energy dissipated \large\rm { = Pt = I^{2} Rt}=Pt=I 2 Rt JoulesConsider Ohm's law \large\rm { V=IR. }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 \large\rm { I²Rt }I²Rt Joules formula.The total current in the circuit is calculated as \large\rm { I = V/R = 6/6 = 1 \ A}I=V/R=6/6=1 A

Energy dissipated \large\rm { = Pt = I^{2} Rt}=Pt=I 2 Rt JoulesConsider Ohm's law \large\rm { V=IR. }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 \large\rm { I²Rt }I²Rt Joules formula.The total current in the circuit is calculated as \large\rm { I = V/R = 6/6 = 1 \ A}I=V/R=6/6=1 ATherefore, the power dissipated P across the 4 ohm resistor for 5 s

Energy dissipated \large\rm { = Pt = I^{2} Rt}=Pt=I 2 Rt JoulesConsider Ohm's law \large\rm { V=IR. }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 \large\rm { I²Rt }I²Rt Joules formula.The total current in the circuit is calculated as \large\rm { I = V/R = 6/6 = 1 \ A}I=V/R=6/6=1 ATherefore, the power dissipated P across the 4 ohm resistor for 5 s\large\rm { = 1^{2} × 4 × 5 = 20 }=1

Energy dissipated \large\rm { = Pt = I^{2} Rt}=Pt=I 2 Rt JoulesConsider Ohm's law \large\rm { V=IR. }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 \large\rm { I²Rt }I²Rt Joules formula.The total current in the circuit is calculated as \large\rm { I = V/R = 6/6 = 1 \ A}I=V/R=6/6=1 ATherefore, the power dissipated P across the 4 ohm resistor for 5 s\large\rm { = 1^{2} × 4 × 5 = 20 }=1 2

Energy dissipated \large\rm { = Pt = I^{2} Rt}=Pt=I 2 Rt JoulesConsider Ohm's law \large\rm { V=IR. }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 \large\rm { I²Rt }I²Rt Joules formula.The total current in the circuit is calculated as \large\rm { I = V/R = 6/6 = 1 \ A}I=V/R=6/6=1 ATherefore, the power dissipated P across the 4 ohm resistor for 5 s\large\rm { = 1^{2} × 4 × 5 = 20 }=1 2 ×4×5=20 Joules

Energy dissipated \large\rm { = Pt = I^{2} Rt}=Pt=I 2 Rt JoulesConsider Ohm's law \large\rm { V=IR. }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 \large\rm { I²Rt }I²Rt Joules formula.The total current in the circuit is calculated as \large\rm { I = V/R = 6/6 = 1 \ A}I=V/R=6/6=1 ATherefore, the power dissipated P across the 4 ohm resistor for 5 s\large\rm { = 1^{2} × 4 × 5 = 20 }=1 2 ×4×5=20 Joulesso, answer is 20 Joules