Question

If two resistors of 47002 and 36002 are connected in series, the effective resistance is given by​

Answer 1

Answer:  83,004 Ω

Let the two resistances be  R1 and R2.

R1 = 47002 Ω

R2 = 36002 Ω

Series connection   Rs=R1+R2

Rs = 47002 + 36002

Rs ( effective resistance ) = 83,004 Ω

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Answer 2

• R_1=47002ohm
• R_2=36002ohm

Both are connected in series

$\boxed{\sf R_{eq}=R_1+R_2}$

$\\ \sf\longmapsto R_{eq}=47002+36002$

$\\ \sf\longmapsto R_{eq}=83004\Omega$

Effective resistance is 83004ohm