The Question is from neet-physics current-electricity , and it was created by the author SUBHANI

Question

Two identical resistors are connected in series across a battery.  If one of the resistors is now short-circuited, the power dissipated in the other resistor will:

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Power dissipation in resistors, series and parallel circuits
 **Related Topics**: Ohm's law, electrical power, circuit analysis
 **Prerequisites**: Understanding of series and parallel resistor combinations, Ohm's law ($V = IR$), and power formula ($P = I^2R = V^2/R$).
## SOLUTION PATHWAY
**Given Information**:
  Two identical resistors, initially in series.
  One resistor is short-circuited.
**Method & Steps**:
 1. **Initial State:** When two identical resistors ($R$) are connected in series across a battery with voltage $V$, the equivalent resistance is $R_{eq} = R + R = 2R$. The current flowing through the circuit is $I = V/R_{eq} = V/(2R)$. The power dissipated in each resistor is $P_1 = I^2R = (V/(2R))^2 R = V^2/(4R)$.
  Reasoning: Using Ohm's law and the power formula.
  Formula/Rule: $V = IR$, $P = I^2R$
 2. **Short-circuiting one resistor:** When one resistor is short-circuited, the current bypasses that resistor. The circuit now effectively consists of only one resistor ($R$) connected to the battery.
  Working: The current through the remaining resistor becomes $I' = V/R$.
  Key Point: The voltage across the remaining resistor is now equal to the battery voltage V.
 3. **Power Dissipation after short circuit:** The power dissipated in the remaining resistor is $P_2 = (I')^2 R = (V/R)^2 R = V^2/R$.
  Result: The power dissipated increased from $V^2/(4R)$ to $V^2/R$, which is a factor of 4.
  Verification: Comparing $P_2$ and $P_1$, we see that $P_2 = 4P_1$.
 ## OPTION ANALYSIS
A :x: Remains the same: Incorrect. Short-circuiting changes the circuit's equivalent resistance and current.
B ✓: Increase by a factor of 4: Correct.  The power dissipated increases by a factor of 4 as shown in the calculations.
C :x: Increase by a factor of 2: Incorrect. The power increases by a factor of 4, not 2.
D :x: Decrease by a factor of 2: Incorrect. The power increases, not decreases.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly applying Ohm's law or power formulas for series and parallel circuits: Ensure you correctly calculate equivalent resistance and current distribution in series and parallel circuits.
  2.Ignoring the effect of short-circuiting: Remember that short-circuiting changes the circuit's configuration and current flow.
 **Quick Tips**:
  1.Draw a circuit diagram to visualize the changes in the circuit.
  2.Remember the power formulas: $P = I^2R = V^2/R = VI$.

SUBHANI · Answer

Remain the same

SUBHANI · Answer

Increase by a factor of 2

SUBHANI · Answer

The correct Option is "Increase by a factor of 4"

SUBHANI · Answer

Decrease by a factor of 2

NEET Physics Current Electricity MCQs

Select Topic

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company