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

In a network of linear resistors fed by ideal voltage sources, the voltage across any resistor is set by a ratio of resistances (a voltage-divider-type expression), not by their absolute size.

1. Halved: Wrong; scaling every resistance by the same factor cancels out of any such ratio.
2. Doubled: Wrong; the source voltage is unchanged, and the ratio is unchanged too.
3. Increased by four times: Wrong; no such compounding occurs from a uniform scale factor.
4. Not changed: Correct, since doubling every resistor leaves every resistance ratio, and therefore every resistor voltage, exactly as before.

Answer 2

Another way to see this is to track what happens to the current and the resistance together, rather than the ratio directly. If every resistance in the network is doubled, then by Ohm's law and the linearity of the whole network, every current in the circuit is exactly halved, since doubling all impedances while keeping the ideal source voltages fixed scales the entire current distribution down by the same factor of one half everywhere.

1. Halved: The voltage across a resistor is (its new current) times (its new resistance), which is \((\tfrac{1}{2}I)\times(2R) = IR\), the original voltage, not half of it, so this option is incorrect.
2. Doubled: By the same computation, \((\tfrac{1}{2}I)\times(2R)=IR\), not \(2IR\), so this option is also incorrect.
3. Increased by four times: The halving of current and doubling of resistance cancel exactly, giving no net multiplication at all, so a four-fold increase does not occur.
4. Not changed: Since the current halves and the resistance doubles, their product, which is the voltage across the resistor, stays exactly the same.

Therefore, the correct answer is not changed.