The given potentiometer has its wire of resistance 10Ω. When the sliding contact is in the middle of the potentiometer wire, the potential drop across 2Ω resistor is :
Applying nodal analysis at point X with potential xV: (x-20)/5 + (x-0)/2 + (x-20)/5 = 0 2 (x-20) + 5x + 2 (x-20) = 0 9x - 80 = 0 => x = 80/9 V Potential drop across 2Ω resistor = x = 80/9 V. (Note: There seems to be a discrepancy in the provided solution and standard circuit analysis. The provided solution calculates the current junction, not a single point.) The solution provided in the image calculates as: (x-0)/5 + (x-20)/5 + (x-20)/2 = 0 2x + 2 (x-20) + 5 (x-20) = 0 9x - 140 = 0 => x = 140/9 V Potential drop across 2Ω = 20 - x = 20 - 140/9 = 40/9 V
<p>Applying nodal analysis at point X with potential xV:<br> (x-20)/5 + (x-0)/2 + (x-20)/5 = 0<br>2 (x-20) + 5x + 2 (x-20) = 0<br>9x - 80 = 0 => x = 80/9 V<br><!- [if !supportLineBreakNewLine]-><!- [endif]->Potential drop across 2Ω resistor = x = 80/9 V.<br><em> (Note: There seems to be a discrepancy in the provided solution and standard circuit analysis. The provided solution calculates the current junction, not a single point.)</em><br>The solution provided in the image calculates as:<br> (x-0)/5 + (x-20)/5 + (x-20)/2 = 0<br>2x + 2 (x-20) + 5 (x-20) = 0<br>9x - 140 = 0<br>=> x = 140/9 V<br>Potential drop across 2Ω = 20 - x = 20 - 140/9 = 40/9 V</p>
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