Question

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Answer 1

answer 32 ohm samjhe ki samjhaye

Answer 2

Answer:

Since two resistor is in series i.e. 100Ω ,  60Ω

$R_{1}$= 100+60= 160Ω

$R_{2\\}$=40Ω

$\frac{1}{R}$=$\frac{1}{160}+\frac{1}{40}$

=$\frac{1+4}{160}$

=5/160

=1/32

R=32Ω

Explanation:

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