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# Engineering

In the figure shown, what is the current (in ampere) drawn from the battery? you are given:

R_{1}=15\Omega ,R_{2}=10\Omega ,R_{3}=20\Omega,R_{4}=5\Omega, R_{5}=25\Omega,R_{6}=30\Omega,E=15V

 

  • Option 1)

    13/24

  • Option 2)

    7/18

  • Option 3)

    9/32

  • Option 4)

    20/3

Answers (1)
S solutionqc

I=\frac{V}{Rer}

R_{eq}=R_{1}+R_{6}+\frac{1}{\frac{1}{R_2}+\frac{1}{R_3+R_4+R_5}}

 

=15+30+\frac{1}{\frac{1}{10}+\frac{1}{20+5+25}}

=15+30+\frac{25}{3}

=\frac{135+25}{3}

=\frac{160}{3}

I=\frac{15}{\frac{160}{3}}=\frac{45}{160}=\frac{9}{32}


Option 1)

13/24

Option 2)

7/18

Option 3)

9/32

Option 4)

20/3

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