Question

Two wires of resistance 2 ohm & 4 ohm are connected in series. The combination is connected to 220 V supply. The power dissipated in 2 ohm resistance is more

Answer 1

Answer:

12100W

Explanation:

Power=V^2/R=220*220/4=12100ohm

Answer 2

Correct Question: Two wires of resistance 2 ohm and 4 ohm are connected in series. The combination is connected to a 220 V supply. The power dissipated in 2 ohm and 4 ohm resistance is equal to?

Answer: Hence the power dissipated by the 2 Ω resistance is equal to 8.07 kW and the power dissipated by the 4 Ω resistance is equal to 16.13 kW.

Given:

Resistance of the first wire (R₁) = 2 Ω

Resistance of the second wire (R₂) = 4 Ω

Supply Voltage (V) = 220 V

To Find:

→ Since the two resistance R₁ and R₂ are connected in series, therefore in order to calculate the net resistance we will add R₁ and R₂.

∴ The net resistance of the circuit (R) = R₁ + R₂ = 2 + 4 = 6 Ω

→ According to Ohm's law the potential difference (V) across a wire is directly proportional to the current (I) flowing through the wire.

                          Voltage (V) ∝ Current (I)

                                          V ∝ I

                                          V = (I) × (R)

                         → where 'R' is the resistance of the wire.

∴ The current (I) flowing through the circuit:

                    $Current(I)=\frac{Voltage(V)}{Resistance(R)}$

                                 $\therefore I=\frac{V}{R} \\\\\therefore I=\frac{220}{6} \\\\\therefore I=\frac{110}{3}\:Ampere$

→ Hence the current through the wires is equal to 110/3 A.

→ The Power (P) dissipated by the wire is given by:

                $Power\:dissipated\:(P)=(V)(I)=I^{2} R=\frac{V^{2} }{R}$

→ The current through the resistors in a series combination remains same.

Hence we can find the Power dissipated (P)  by the wires by using the formula: P = I²R

→ Power dissipated (P₁) by the  resistor: (R₁):

       $\therefore P_{1}=I^{2} R_{1} \\\\\therefore P_{1}= (\frac{110}{3} )^{2} \times(2)\\\\\therefore P_{1} =\frac{24200}{3} \:Watt=8.07\:kW$

→ Power dissipated (P₂) by the  resistor: (R₂):

       $\therefore P_{2}=I^{2} R_{2} \\\\\therefore P_{2} =(\frac{110}{3} )^{2} \times(4)\\\\\therefore P_{2} =\frac{24200}{3} \:Watt=16.13\:kW$

Hence the power dissipated by the 2 Ω resistance is equal to 8.07 kW and the power dissipated by the 4 Ω resistance is equal to 16.13 kW.

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