Question-216016

Question Number 216016 by Tawa11 last updated on 25/Jan/25

Answered by som(math1967) last updated on 25/Jan/25

c u r r e n t 20 Ω w h e n b o t h s w i t c h

o p p e n = \frac{6}{80} = \frac{3}{40} a m p

c u r r e n t f l o w c c t w h e n s_{1} s_{2} b o t h

c l o s e 6 \div (50 + \frac{20 \times R}{20 + R})

= 6 \div (\frac{1000 + 70 R}{20 + R})

= \frac{6 (20 + R)}{(1000 + 70 R)}

p d a t 50 Ω \frac{300 (20 + R)}{10 (100 + 7 R)}

p d a t 20 Ω 6 - \frac{30 (20 + R)}{100 + 7 R}

∴ 20 \times \frac{3}{40} = 6 - \frac{30 (20 + R)}{100 + 7 R}

\Rightarrow \frac{30 (20 + R)}{100 + 7 R} = 6 - \frac{3}{2}

\Rightarrow \frac{30 (20 + R)}{100 + 7 R} = \frac{9}{2}

\Rightarrow 400 + 20 R = 300 + 21 R

∴ R = 100 Ω

Commented by Tawa11 last updated on 25/Jan/25

God bless you sir .

I really appreciate .

Commented by Tawa11 last updated on 28/Jan/25

Sir, please help .

Q 216153

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