The Voltage Divider Problem

This diagram shows a network of resistors connected up to one battery. For a series circuit, the voltage of the battery is "divided up" into parts: here the battery voltage E is split up amongst parts we can call V₁, V_AB, V₄, V₅ and V₆. Each of the parts is smaller than the whole. The larger voltages will be found across the larger resistances (that is where the energy loss is greatest as charges pass through).

The network also contains two resistors in parallel. For these, the currents are additive and each of the currents (I₂ and I₃) is smaller than the amount that entered the junction before the split (I₁). The voltage V_AB is common to both resistors, and determines the currents through them: whichever resistor has the least resistance will carry the larger of the two currents.

The two resistors R₃ and R₆ shown with arrows through them have variable values. Suppose you are told to figure out what happens to the current through resistor R₂ if either R₃ or R₆ are increased (one or the other). How do you proceed?

First, let's consider the case where R₃ is increased. You might begin by thinking: the greatest amount of current takes the path of least resistance, and quickly conclude that current will be diverted away from R₃, resulting in an increase in current through R₂. However, there is a flaw in the conclusion: you have assumed that the current I₁ coming into junction A in the picture is staying constant; it is not.

To draw a reliable conclusion about the current through R₂, you have to consider the voltage V_AB across that resistor. Because R₃ is already connected (it is not being added to the circuit by closing a switch; that is a different case), the combined resistance R₂₃ increases when R₃ is increased. When the series nature of the circuit is then considered, there is more resistance between points A and B so more of the battery's divided voltage will now appear there compared with before. Since V_AB increases, it follows that I₂ increases. The current I₁ actually decreases, since the resistance of the entire network has increased (so it is inaccurate to just claim that R₂ steals current away from R₃).

Second, let's consider the case where R₆ is increased (R₃ unchanged). Since R₆ is in series with the rest of the network, an increase in R₆ leads to an increase in the network resistance. The current through R₆ is the same as the battery current (also the same as the current through R₁), so it will decrease. This decreased current arrives at junction A and splits up into two parts. Assuming that the two paths maintain their relative resistances, both those current parts will decrease. Hence we conclude that I₂ decreases

We can answer this second question also by considering the concept of the "voltage divider" or series circuit. Since R₆ increases, the portion of the battery voltage which is found across it also increases (more energy loss). This leaves less voltage available to be divided up amongst the other series components, including the part between junctions A and B. When V₆ increases, this causes V₁, V_AB, V₄ and V₅ to all decrease. Since V_AB decreases, it follows that I₂ decreases.

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