MHT CET · Physics · Current Electricity
Five current carrying conductors meet at a point ' \(\mathrm{O}\) ' as shown in figure. The magnitude and direction of the current in conductor ' \(\mathrm{OP}\) ' is
- A \(6.5 \mathrm{~A}\) from \(\mathrm{O}\) to \(\mathrm{P}\).
- B \(9 \mathrm{~A}\) from \(\mathrm{P}\) to \(\mathrm{O}\).
- C \(10.5 \mathrm{~A}\) from \(\mathrm{P}\) to \(\mathrm{O}\).
- D \(11.5 \mathrm{~A}\) from \(\mathrm{O}\) to \(\mathrm{P}\).
Answer & Solution
Correct Answer
(D) \(11.5 \mathrm{~A}\) from \(\mathrm{O}\) to \(\mathrm{P}\).
Step-by-step Solution
Detailed explanation
Using Kirchhoff's current Law, Current flowing in = Current flowing out
\(\begin{aligned}
& 10+2.5+5=6+\mathrm{x} \\
\therefore \mathrm{x} & =17.5-6 \\
& =11.5 \mathrm{~A}
\end{aligned}\)
\(\therefore\) Magnitude and direction of the current in 'OP' should be \(11.5 \mathrm{~A}\) from \(\mathrm{O}\) to \(\mathrm{P}\).
\(\begin{aligned}
& 10+2.5+5=6+\mathrm{x} \\
\therefore \mathrm{x} & =17.5-6 \\
& =11.5 \mathrm{~A}
\end{aligned}\)
\(\therefore\) Magnitude and direction of the current in 'OP' should be \(11.5 \mathrm{~A}\) from \(\mathrm{O}\) to \(\mathrm{P}\).
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