Physics, asked by chocolatelover01, 7 months ago


An electric lamp of 100 ohms, toaster of resistance 50 ohms and a water filter of resistance 500ohms are connecte in parallel to a 220 V source. What is the resistance of an electric iron connected to the same sources takes as much current as all three appliances, and what is the current through it ?​

Answers

Answered by mbakshi37
0

Explanation:

1. in parallel i/ R is summ of All 1/Rs , so effective R is:

500 /16 ohms.

2. I = V/ R = 220×16/500=: ? 4.5 amps

Answered by TheValkyrie
6

Answer:

\bigstar{\bold{Resistance=31.25\: \Omega}}

\bigstar{\bold{Current=7.04\:A}}

Explanation:

\Large{\underline{\underline{\sf{Given:}}}}

  • Resistance of electric lamp = 100 Ω
  • Resistance of toaster = 50 Ω
  • Resistance of water filter = 500 Ω
  • Voltage (V) = 220 V
  • Appliances are connected in parallel

\Large{\underline{\underline{\sf{To\:Find:}}}}

  • Resistance of the electric iron
  • Current through the electric iron if it takes as much current as the other three appliances.

\Large{\underline{\underline{\sf{Solution:}}}}

→ Since the appliances are connected in parallel, we have to find the total resistance.

→ The total resistance when resistors are connected in parallel is given by,

  \dfrac{1}{R} =\dfrac{1}{R_1} +\dfrac{1}{R_2} +\dfrac{1}{R_3}

→ Substituting the individual resistances,

  \dfrac{1}{R}=\dfrac{1}{100} +\dfrac{1}{50}+\dfrac{1}{500}

 \dfrac{1}{R}=\dfrac{5}{500} +\dfrac{10}{500}+\dfrac{1}{500}

 \dfrac{1}{R}=\dfrac{16}{500}

 R = 500/16 = 31.25 Ω

→ Hence resistance of the electric iron is 31.25 Ω

\boxed{\bold{Resistance=31.25\: \Omega}}

→ By Ohm's law we know that,

 I = V/R

→ Substituting the datas we get the value of current (I)

  I = 220/31.25

  I = 7.04 A

→ Hence current flowing through the electric iron is 7.04 A

\boxed{\bold{Current=7.04\:A}}

\Large{\underline{\underline{\sf{Notes:}}}}

→ The total resistance when resistors are connected in series is given by,

   R = R_1+R_2+R_3+.....R_n

→ The total resistance when resistors are connected in parallel is given by,

   \dfrac{1}{R} =\dfrac{1}{R_1} +\dfrac{1}{R_2} +\dfrac{1}{R_3}+.....+\dfrac{1}{R_n}

 

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