Physics, asked by rizu890, 4 months ago

three wires of resistances 20, 24, and 30 ohm are arranged in
1)series
2)parallel arrangements. Find the equivalent resistance of the circuit in each situation

Answers

Answered by ItzSuperBranded03
0

1. Resistance ( in series),

R = R1 +R2+R3

=> 20+24+30

=> 64 ohm

\huge\mathcal{\fcolorbox{lime}{black}{\pink{From:-ItzSuperBranded03}}}

Answered by Anonymous
4

GIVEN:

\sf{R_1=20\Omega}

\sf{R_2=24\Omega}

 \sf{R_3=30\Omega}

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TO FIND:

\sf{R_{(Parallel)}=?}

\sf{R_{(Series)}=?}

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SOLUTION:

   {\large{\underline{ \underline{ \mathbb{SERIES}}}}}

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{\sf{R_{(Series)}=R_1+R_2+R_3}}

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{\sf{R_{(Series)}=24\Omega+20\Omega+30\Omega}}

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{\sf{R_{(Series)}=74\Omega}}

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  {\large{\underline{ \underline{ \mathbb{PARALLEL}}}}}

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{ \sf{ \dfrac{1}{R_{(Parallel)}}= \dfrac{1}{R_1}+ \dfrac{1}{R_2}+ \dfrac{1}{R_3}}}

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{ \sf{ \dfrac{1}{R_{(Parallel)}}= \dfrac{1}{20}+ \dfrac{1}{24}+ \dfrac{1}{30}}}

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{ \sf{ \dfrac{1}{R_{(Parallel)}}= \dfrac{6 + 5 + 4}{120} }}

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{ \sf{ \dfrac{1}{R_{(Parallel)}}= \dfrac{15}{120} }}

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{ \sf{ \dfrac{1}{R_{(Parallel)}}= \dfrac{1}{8} }}

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{ \sf{R_{(Parallel)}= {8 \Omega} }}

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∴The answers are:-

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 \sf{R_{(Parallel)}=8\Omega}

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\sf{R_{(Series)}=74\Omega}

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{ \bf {\large {\underline{ \underline{Additional \:  Information: }}}}}

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Resistance is always maximum in series whereas lowest in parallel connection.

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Ammeter is a device used to measure current in a circuit.

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Ammeter is always connected in series connection in circuit.

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Voltmeter is a device used to measure potential difference (or voltage) in a circuit.

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Voltmeter is always connected in parallel connection with circuit.

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Kindly see this answer on web for better understanding. ‎ ‎

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