MHT CET · Physics · Alternating Current
A circular coil of radius 'R' has a resistance of \(40 \Omega\). Figure shows two points 'P' and 'Q' on the circumference separated by a distance \(\frac{\pi \mathrm{R}}{2}\), which are connected to a \(16 \mathrm{~V}\) battery with internal resistance of \(0.5 \Omega\). What is the value of current 'I' flowing through the circuit?
- A 1A
- B 0.5A
- C 3A
- D 2A
Answer & Solution
Correct Answer
(D) 2A
Step-by-step Solution
Detailed explanation
The total resistance between points \(\mathrm{P}\) and \(\mathrm{Q}\).
\(\begin{array}{l}
\mathrm{RPQ}=(30 \| 10+0.5) \\
=\frac{30 \times 10}{30+10}+0.5=8 \Omega
\end{array}\)
The current flowing through the circuit,
\(\begin{array}{l}
V=I R_{P Q} \\
\Rightarrow I=\frac{V}{R_{P Q}}
\end{array}\)
where, \(V\) is the voltage of source.
\(\therefore \mathrm{I}=\frac{16 \mathrm{~V}}{8 \Omega}=2 \mathrm{~A}\)
\(\begin{array}{l}
\mathrm{RPQ}=(30 \| 10+0.5) \\
=\frac{30 \times 10}{30+10}+0.5=8 \Omega
\end{array}\)
The current flowing through the circuit,
\(\begin{array}{l}
V=I R_{P Q} \\
\Rightarrow I=\frac{V}{R_{P Q}}
\end{array}\)
where, \(V\) is the voltage of source.
\(\therefore \mathrm{I}=\frac{16 \mathrm{~V}}{8 \Omega}=2 \mathrm{~A}\)
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