JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
In the figure shown below, a resistance of \(150.4 \Omega\) is connected in series to an ammeter A of resistance \(240 \Omega\). A shunt resistance of \(10 \Omega\) is connected in parallel with the ammeter. The reading of the ammeter is _____ mA.
- A 10
- B 15
- C 20
- D 5
Answer & Solution
Correct Answer
(D) 5
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2 \\ & \mathrm{R}_{\mathrm{eq}}=150.4+\frac{240 \times 10}{250} \\ & =150.4+9.6=160 \Omega \\ & \mathrm{I}_1=\frac{\mathrm{I} \mathrm{R}_2}{240} \\ & \mathrm{I}_1=\frac{\mathrm{I} \times 9.6}{240} \\ &…
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