Question

A network contains linear resistors and ideal voltage sources. If values of all the resistors are doubled, then the voltage across each resistor is

• A. halved
• B. doubled
• C. increased by four times
• D. not changed

Hint:

In linear circuits with ideal voltage sources, scaling all resistances equally does not change voltage distribution—only currents change.

Answer 1

The Correct Option is D

Step 1: Understanding the given network.
The network consists of linear resistors and ideal voltage sources. Ideal voltage sources maintain a fixed voltage regardless of the current drawn from them.
Step 2: Effect of doubling resistances.
When all resistances in the circuit are doubled, the total resistance of the circuit increases. As a result, the current flowing in the circuit decreases according to Ohm’s law.
Step 3: Voltage across each resistor.
Although the current decreases, the voltage distribution in a linear resistive network with ideal voltage sources depends only on the ratio of resistances, not their absolute values. Since all resistances are scaled by the same factor, their ratios remain unchanged.
Step 4: Analyzing the options.
(A) Voltage is not halved because voltage division ratios remain the same.
(B) Voltage does not double since source voltages are unchanged.
(C) Voltage cannot increase four times without change in source voltage.
(D) Correct — the voltage across each resistor remains unchanged.
Step 5: Final conclusion.
Thus, even after doubling all resistances, the voltage across each resistor remains the same.

Answer 2

A concrete example makes this easy to check. Suppose an ideal voltage source of value \(V\) feeds two resistors \(R_1\) and \(R_2\) in series, so the voltage across \(R_2\) is \(V\dfrac{R_2}{R_1+R_2}\) by the voltage-divider rule. Doubling both resistors to \(2R_1\) and \(2R_2\) gives a new voltage across the second resistor of \(V\dfrac{2R_2}{2R_1+2R_2} = V\dfrac{R_2}{R_1+R_2}\), which is algebraically identical to the original expression, since the factor of 2 cancels top and bottom.

1. Halved: This would require the ratio \(R_2/(R_1+R_2)\) itself to be halved by the scaling, but multiplying every resistor by the same constant leaves every such ratio unchanged.
2. Doubled: This would require the source voltage or the ratio to double, but the source voltage is set independently by the ideal source and does not change just because resistances change.
3. Increased by four times: This large a change is not produced by scaling all resistances uniformly, since the scaling factor cancels exactly in the divider ratio rather than compounding.
4. Not changed: This matches the algebraic result exactly: uniformly scaling every resistor leaves every voltage-divider ratio, and hence every resistor's voltage, unchanged.

Because the network contains only linear resistors and ideal voltage sources, scaling every resistance by the same factor scales every current inversely but leaves every voltage division ratio, and therefore every voltage across a resistor, exactly as it was.

Therefore, the correct answer is not changed.