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GUJCET · Physics · Current Electricity
A wire is bent in the form of circle of radius 2 m . Resistance per unit length of wire is \(\frac{1}{\pi} \frac{\Omega}{m}\). Battery of 6 V is connected between A and B. \(\angle \mathrm{AOB}=90^{\circ}\). Find the current through the battery.
- A 8 A
- B 4 A
- C 3 A
- D 9 A
Answer & Solution
Correct Answer
(A) 8 A
Step-by-step Solution
Detailed explanation
(A)
length of wire \(=2 \pi r\)
\(\begin{array}{l}=2 \pi(2) \\ =4 \pi m\end{array}\)
resistance of wire \(=\frac{1}{\pi} \times 4 \pi\)
\(=4 \Omega\)
minor arc AB resistance \(=1 \Omega\)
major arc AB resistance \(=3 \Omega\)
both section are connected in parallel,
\(\begin{array}{l} R _{ p }=\frac{1 \times 3}{1+3} \\ R _{ p }=\frac{3}{4} \Omega\end{array}\)
current flowing the battery
\(\begin{aligned} I =\frac{\varepsilon}{ R _{ p }} \\ \therefore I =\frac{6}{\frac{3}{4}} \\ \therefore I =\frac{6 \times 4}{3} \\ \therefore I =8 A\end{aligned}\)
length of wire \(=2 \pi r\)
\(\begin{array}{l}=2 \pi(2) \\ =4 \pi m\end{array}\)
resistance of wire \(=\frac{1}{\pi} \times 4 \pi\)
\(=4 \Omega\)
minor arc AB resistance \(=1 \Omega\)
major arc AB resistance \(=3 \Omega\)
both section are connected in parallel,
\(\begin{array}{l} R _{ p }=\frac{1 \times 3}{1+3} \\ R _{ p }=\frac{3}{4} \Omega\end{array}\)
current flowing the battery
\(\begin{aligned} I =\frac{\varepsilon}{ R _{ p }} \\ \therefore I =\frac{6}{\frac{3}{4}} \\ \therefore I =\frac{6 \times 4}{3} \\ \therefore I =8 A\end{aligned}\)
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