KCET · Physics · Current Electricity
The equivalent resistance between the points \(A\) and \(B\) in the following circuit is
- A \(0.5 \Omega\)
- B \(5.5 \Omega\)
- C \(0.05 \Omega\)
- D \(5 \Omega\)
Answer & Solution
Correct Answer
(B) \(5.5 \Omega\)
Step-by-step Solution
Detailed explanation
Let \(x\) be the equivalent resistance of the
circuit, then
\(\therefore R_{A B}=x \)
\( \Rightarrow \frac{x \times 2}{2+x}+2+2=x \)
\( \Rightarrow \frac{2 x}{x+2}=x-4 \)
\( \Rightarrow 2 x=(x-4)(x+2) \)
\( \Rightarrow 2 x=x^2+2 x-4 x-8 \)
\( \Rightarrow x^2-4 x-8=0 \)
\( \Rightarrow \frac{-(-4) \pm \sqrt{(-4)^2-4 \times 1 \times(-8)}}{2 \times 1} \)
\( =\frac{4 \pm \sqrt{48}}{2}=\frac{4 \pm 4 \sqrt{3}}{2} \)
\( =2+2 \sqrt{3}=5.46 \Omega \approx 5.5 \Omega\)
circuit, then
\(\therefore R_{A B}=x \)
\( \Rightarrow \frac{x \times 2}{2+x}+2+2=x \)
\( \Rightarrow \frac{2 x}{x+2}=x-4 \)
\( \Rightarrow 2 x=(x-4)(x+2) \)
\( \Rightarrow 2 x=x^2+2 x-4 x-8 \)
\( \Rightarrow x^2-4 x-8=0 \)
\( \Rightarrow \frac{-(-4) \pm \sqrt{(-4)^2-4 \times 1 \times(-8)}}{2 \times 1} \)
\( =\frac{4 \pm \sqrt{48}}{2}=\frac{4 \pm 4 \sqrt{3}}{2} \)
\( =2+2 \sqrt{3}=5.46 \Omega \approx 5.5 \Omega\)
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