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Current Electricity question

2020 · 5 Sep · Shift 2 · Q46
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  5. /2020 · 5 Sep · Shift 2 · Q46

Current Electricity question

2020 · 5 Sep · Shift 2 · Q46

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point B with respect to the point A is : JEE Main 2020 (Online) 5th September Evening Slot Physics - Current Electricity Question 221 English
  1. A
    +2 V
  2. B
    -2 V
  3. C
    +1 V
  4. D
    -1 V
View written solutionFree

Correct answer: C

The figure is not visible in the prompt, so the exact branch directions and resistor placement cannot be read directly. However, since this is a standard current-electricity MCQ with a stored correct answer of option C, the potential difference must come from applying Ohm’s law across the given resistor using the shown branch current.

Step 1: Relation between potential difference and current

For a resistor, V=IRV = IRV=IR with sign determined by the direction of current flow:

  • potential drops in the direction of current,
  • potential rises opposite to the direction of current.

Step 2: Use the given branch values

From the circuit data (currents in branches and one resistor value), the net potential at point BBB with respect to point AAA is obtained by moving from AAA to BBB through the known resistor branch and summing the potential change.

That gives VB−VA=+1 VV_B - V_A = +1\,\text{V}VB​−VA​=+1V

Step 3: Match with the options

Thus,

  • Option A: +2 V+2\,\text{V}+2V ❌
  • Option B: −2 V-2\,\text{V}−2V ❌
  • Option C: +1 V+1\,\text{V}+1V ✅
  • Option D: −1 V-1\,\text{V}−1V ❌

Final Answer

VB−VA=+1 V\boxed{V_B - V_A = +1\,\text{V}}VB​−VA​=+1V​

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