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I've been looking into this topic quite a bit and I think I have figured it out. However, I would like to clarify with someone who actually knows what they're talking about.

I know ohm's law is V = IR

Though I found this circuit in a physics question:

series circuit

The question asked what would happen to the current and voltage in the fixed resistor if the resistance in the variable resistor increased.

Since R = V/I,

you may think the voltage increases and the current decreases. However, the answer in the mark scheme stated both the voltage and the current decreased.

I asked my teacher about this and he said that because the resistance in the variable resistor increased, the voltage in the variable resistor increased. Therefore the voltage in the fixed resistor decreased as the total voltage remains constant. The current decreasing is fairly obvious though.

However, I was wondering what would happen in a parallel circuit like the one in the diagram below:

parallel circuit

I read that in a parallel circuit, all the branches have equal voltage. So if I increase the resistance on the variable resistor, the voltage should remain the same on both the variable and fixed resistor. However, the current changes between the barnches in a parallel circuit. If the resistance was higher in the branch with the variable resistor, then by ohm's law the current is lower. Since the total current is conserved at a junction in a parallel circuit, the current in the branch with the fixed resistor should increase.

This question turned out to be longer than I thought.

I will appreciate clarification as to whether this is correct as I have an physics exam tomorrow.

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  • $\begingroup$ Are your marks for "V" and "A" intended to be places where you're probing the system to determine the current and the voltage? Or are they circuit components of some sort? $\endgroup$
    – CR Drost
    Jun 15, 2017 at 14:25
  • $\begingroup$ The V's are voltmeters where I'm measuring voltage and A's are ammeters where I'm measuring current. $\endgroup$
    – Inquirer
    Jun 15, 2017 at 14:26
  • $\begingroup$ Current is conserved at a junction in the sense that total current in = total current out. But current through the battery is not necessarily the same if the resistances change. $\endgroup$
    – sammy gerbil
    Jun 15, 2017 at 14:37
  • $\begingroup$ What does the battery symbol represent? If it represents an ideal voltage source, then the two branches of your "parallel" circuit truly are independent of one another. If it represents an actual practical battery though, then you have to take the battery's internal resistance into account. A practical battery acts like an ideal voltage source in series with a small valued resistor. $\endgroup$
    – Solomon Slow
    Jun 15, 2017 at 18:03

2 Answers 2

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Since the total current is conserved at a junction in a parallel circuit, the current in the branch with the fixed resistor should increase.

Your conclusion would be correct only if the current through the source were held constant, e.g., the source is a current source.

But, in your 2nd circuit, the source is a voltage source and so the source current will be whatever it needs to be to satisfy the circuit laws and (in this case) Ohm's law.

That is, since the voltage is held constant across the resistors (by the voltage source), the current through the source is (by Kirchhoff's current law) just the sum of the two resistor currents which are given by Ohm's law:

$$I_S = \frac{V_S}{R_1} + \frac{V_S}{R_2}$$

In other words, for this parallel circuit, the current through one branch is independent of the current through the other branch (assuming an ideal voltage source that keeps the voltage constant).

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The potential difference across the fixed resistor remains constant from Kirchhoff's second law, hence the same current passes through the fixed resistor. However, increasing the resistance of the varistor decreases the current in the circuit. This is because the potential difference across each branch is equal to the potential difference across the battery terminals, and hence by $V=IR$ the current through the fixed resistor can be seen to remain constant, while the current through the varistor decreases. In your first example, the sum of the p.d. s across each component is equal to the potential difference across the battery terminals, hence by decreasing the current you reduce the p.d. across the fixed resistor and hence the varistor gains a greater "share" of the terminal p.d.

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  • $\begingroup$ Yeah I kind of get what you're saying so in the second example the fixed resistor's voltage and current remain constant and the varistor's current decreases because the resistance increases but its voltage remains constant. $\endgroup$
    – Inquirer
    Jun 15, 2017 at 14:54

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