The Question is from neet-physics current-electricity ohms-law-combination-of-resistors, and it was created by the author NAGA SAI

Question

A network of resistors is connected to a battery of emf $E$ as shown.  If the internal resistance of the battery is negligible, the potential difference across the resistor $R_1$ is:

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Ohm's Law and Circuit Analysis (Series and Parallel Resistors)
	**Related Topics**: Equivalent Resistance, Voltage Division
	**Prerequisites**: Understanding of series and parallel resistor combinations, Ohm's Law ($V = IR$)
## SOLUTION PATHWAY
**Given Information**:
	- Battery emf: $E$
	- Negligible internal resistance
	- Resistors: $R_1$, $R_2$, $R_3$, $R_4$, $R_5$
**Method & Steps**:
	1. **Simplify the circuit:**
		Reasoning: Combine resistors in series and parallel to find the equivalent resistance of the network.
		Formula/Rule: Series: $R_{eq} = R_1 + R_2 + ...$
			Parallel: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ...$
	2. **Calculate the equivalent resistance of $R_2$, $R_3$: and $R_4$, $R_5$**
		Working: $R_{23} = R_2 + R_3$ (Series)
			$R_{45} = R_4 + R_5$ (Series)
		Key Point: These combined resistances are now in parallel.
	3. **Calculate the equivalent resistance of $R_{23}$ and $R_{45}$**
		Working: $\frac{1}{R_{2345}} = \frac{1}{R_{23}} + \frac{1}{R_{45}} \implies R_{2345} = \frac{R_{23}R_{45}}{R_{23} + R_{45}} = \frac{(R_2 + R_3)(R_4 + R_5)}{(R_2 + R_3) + (R_4 + R_5)}$
	4. **Calculate total equivalent resistance:**
		Working: $R_{eq} = R_1 + R_{2345} = R_1 + \frac{(R_2 + R_3)(R_4 + R_5)}{(R_2 + R_3) + (R_4 + R_5)}$
	5. **Apply Voltage Division:**
		Working: $V_{R_1} = E * \frac{R_1}{R_{eq}} = E * \frac{R_1}{R_1 + \frac{(R_2 + R_3)(R_4 + R_5)}{(R_2 + R_3) + (R_4 + R_5)}} = E*\frac{R_1(R_2+R_3+R_4+R_5)}{R_1(R_2+R_3+R_4+R_5)+(R_2+R_3)(R_4+R_5)}$
		Result: $V_{R_1} = E*\frac{R_1(R_2+R_3+R_4+R_5)}{R_1(R_2+R_3+R_4+R_5)+(R_2+R_3)(R_4+R_5)}$
		Verification: The voltage across $R_1$ must be less than $E$.
## OPTION ANALYSIS
[A] :x: This formula doesn't account for the parallel combination of resistors.
[B] ✓: This is the correct application of the voltage divider rule after simplifying the circuit.
[C] :x: This would only be true if $R_1$ was the only resistor in the circuit.
[D] :x: This formula only considers $R_1$ and $R_2$ and ignores the other resistors.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly combining resistors: Ensure you correctly identify series and parallel combinations.
		2. Forgetting voltage division: Remember that the voltage across a resistor in a series circuit is proportional to its resistance.
	**Quick Tips**:
		1. Redraw the circuit after each simplification step to visualize the equivalent resistance.
		2. Double-check your calculations to avoid algebraic errors.

NAGA SAI · Answer

E*R1/(R1+R2+R3+R4+R5)

NAGA SAI · Answer

The correct Option is "E*R1(R2+R3+R4+R5)/(R1(R2+R3+R4+R5)+(R2+R3)(R4+R5))"

NAGA SAI · Answer

E

NAGA SAI · Answer

E*R1/(R1+R2)

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