JEE Mains · Physics · STD 12 - 3. current electricity
Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance \(R_p=1 \Omega\) as shown in the figure. An external resistance of \(R_e=2 \Omega\) is connected via the sliding contact. The electric current in the circuit is :
- A 0.9 A
- B 1.35 A
- C 0.3 A
- D 1.0 A
Answer & Solution
Correct Answer
(D) 1.0 A
Step-by-step Solution
Detailed explanation
The circuit can be considered as \(\begin{aligned} & \therefore \mathrm{R}_{\text {eq }}=0.5+\frac{0.5 \times 2}{2+0.5}=\left(\frac{5}{10}+\frac{10}{25}\right) \Omega \\ & =\frac{45}{50}=\frac{9}{10}=0.9 \\ & \therefore \mathrm{i}=\frac{0.9}{0.9}=1 \mathrm{~A}\end{aligned}\)
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