JEE Mains · Physics · STD 12 - 3. current electricity
In the given figure, there is a circuit of potentiometer of length \(A B=10 \,{m}\). The resistance per unit length is \(0.1 \,\Omega\) per \({cm}\). Across \({AB}\), a battery of emf \({E}\) and internal resistance ' \({r}^{\prime}\) is connected. The maximum value of emf measured by this potentiometer is : (In \(V\))
- A \(6\)
- B \(2.25\)
- C \(5\)
- D \(2.75\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
Max. voltage that can be measured by this potentiometer will be equal to potential drop across AB \(R _{ AB }=10 \times 0.1 \times 100=100 \Omega\) \(\therefore V _{ AB }=\frac{6}{20+100} \times 100=6 \times \frac{100}{120}=5 V\)
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