NEET · Physics · STD 12 - 3. current electricity
The sliding contact \(C\) is at one fourth of the length of the potentiometer wire \(( AB )\) from \(A\) as shown in the circuit diagram. If the resistance of the wire \(AB\) is \(R _0\), then the potential drop \(( V )\) across the resistor \(R\) is
- A \(\frac{4 V _0 R }{3 R _0+16 R }\)
- B \(\frac{4 V _0 R }{3 R _0+ R }\)
- C \(\frac{2 V _0 R }{4 R _0+ R }\)
- D \(\frac{2 V _0 R }{2 R _0+3 R }\)
Answer & Solution
Correct Answer
(A) \(\frac{4 V _0 R }{3 R _0+16 R }\)
Step-by-step Solution
Detailed explanation
In series, potential divides in direct ratio of resistance, So, \(V_{A C}=\frac{R_{A C}}{R_{A C}+R_{C B}} V_0\)
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