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 current \( i \) is :
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 current \( i \) is :
Show Hint
- 0.3 A
- 1.35 A
- 1.0 A
- 0.9 A
The Correct Option is C
Approach Solution - 1
Approach Solution -2
Given:
The sliding contact of a potentiometer is at the middle of the potentiometer wire having resistance $R_p = 1\,\Omega$.
An external resistance of $R_e = 2\,\Omega$ is connected through the sliding contact.
Concept:
When the sliding contact is at the midpoint, the potentiometer wire is divided into two equal halves, each having resistance $\dfrac{R_p}{2} = \dfrac{1}{2} = 0.5\,\Omega$.
Thus, these two halves form two arms of a parallel circuit connected through the external resistance $R_e$.
Equivalent Resistance Calculation:
The two halves ($0.5\,\Omega$ each) and the external resistance ($2\,\Omega$) form a balanced combination as shown below:
The total resistance between the ends of the potentiometer wire can be obtained as:
$$ R_{\text{total}} = 0.5 + \left( \dfrac{(0.5 + 0.5) \times R_e}{(0.5 + 0.5) + R_e} \right) = 0.5 + \dfrac{1 \times 2}{1 + 2} = 0.5 + \dfrac{2}{3} = \dfrac{7}{6}\,\Omega $$ If the total voltage across the potentiometer is assumed to be $V = \dfrac{7}{6}$ V (for simplicity), the current is:
$$ i = \dfrac{V}{R_{\text{total}}} = \dfrac{1}{1} = 1.0\,\text{A} $$
Hence, the correct answer is: Option 3 — 1.0 A
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