Physics, asked by Hardik3803, 11 months ago

35: a hot plate of an electric oven connected to 220 v line has two resistance coils a and b, each of 24 q resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answers

Answered by LovelyG
107

Answer:

\large{\underline{\boxed{\sf (i)\: 9.166 A}}}

\large{\underline{\boxed{\sf (ii)\: 4.58A}}}

\large{\underline{\boxed{\sf  (iii)\: 18.33 A}}}

Explanation:

Given that ;

Supply voltage, V = 220 V

Resistance of one coil, R =24 Ω

________________________

  • Case I :

(i) Coils are used separately,

According to Ohm’s law,

\text V = \text{I}_1 \text{R}_1

\text{I}_1 is the current flowing through the coil.

\text{I}_1 = \frac{\text{V}}{\text{R}_1}\\\\\frac{220}{24} = 9.166 \text{A}

Hence, when coils are used separately, the current flow is 9.166 A.

____________________________

  • Case II :

(ii) Coils are connected in series,

R2=24 Ω+24 Ω=48 Ω

According to Ohm’s law;

\text{I}_2 = \frac{\text{V}}{\text{R}_2}\\\\\frac{220}{48} = 4.58 \text{A}

Hence, the current flowing through the series circuit is 4.58A.

____________________________

  • Case III :

(iii) Coils are connected in parallel,

\frac{1}{\text{R}_3} = \frac{1}{24}+\frac{1}{24}\\\\\frac{1}{\text{R}_3} = \frac{1+1}{24}\\\\\frac{1}{\text{R}_3} =\frac{2}{24}\\\\\frac{1}{\text{R}_3} =\frac{1}{12}\\\\\implies \text{R}_3=12 \ \Omega

Now, using ohm's law-

\text{I}_2 = \frac{\text{V}}{\text{R}_2}\\\\\frac{220}{12} = 18.33 \text{A}

Hence, the current flowing through the parallel circuit is 18.33A.

Answered by Anonymous
153

\mathabb{\huge{\boxed{Answer:-\\ \\(i)9.166A\\ \\(ii) 4.58A\\(iii) 18.33A}}}

Step by step explanation :-

Voltage (V) = 220v

Resistance of one coil = 24 ohm

______________________________

(i):- Coils are used separately :-

According to ohm's law, V=IR

I1 is the current following through coil

I1= V/ R1 = 220/24 =9.166A

______________________________

(ii) Coils are connected in series :-

R2 = 24+24 = 48 ohm

According to ohm's law

I2 = V/R2 = 220/48 = 4.58 A

____________________________

(iii) Coils are connected in parallel :-

1/R3 = 1/24+1/24

1/R3 = 2/24( By taking LCM)

1/R3 = 1/12

1/R3 = 12ohm

according to ohm's law

I3 = V/R3 = 220/12

=18.33A

_____________________________

:-)

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