In a parallel circuit, every branch has the same potential difference as the supply. When one branch opens, the other branches keep their current, and only the total current from the cell falls.
This lesson belongs to electrical reasoning from diagrams and measurements. The next lesson, comparing potential difference readings, keeps the same circuit and asks what each voltmeter shows.
Worked example: two lamps in parallel
An original circuit: a 6 V supply is connected to lamps L1 and L2 in parallel. Each lamp has a resistance of 12 Ω. Lamp L2 is then removed, leaving its branch open. Assume the supply is ideal.
| Quantity | Before | After | Reason |
|---|---|---|---|
| p.d. across L1 | 6 V | 6 V | Still connected directly to the supply |
| Current through L1 | 0.5 A | 0.5 A | I = V ÷ R with V and R unchanged |
| Brightness of L1 | normal | normal | Same current, same power |
| Current in L2 branch | 0.5 A | 0 A | Branch is open |
| Total current from supply | 1.0 A | 0.5 A | Sum of branch currents |
A full answer states what stays the same, and then gives the reason: the p.d. across L1 has not changed, so its current has not changed.
Why does the series case differ?
In a series circuit, there is one path. If one lamp is removed, the path breaks and the current in every part falls to zero, so all lamps go out.
This contrast is the core of the topic. In series, the current is shared. In parallel, the potential difference is shared.
The mistake that costs marks
A common slip is to say the remaining lamp gets brighter because the current has nowhere else to go.
| Claim | Why it fails |
|---|---|
| “Current has nowhere else to go, so L1 is brighter” | A parallel branch takes its own current from the supply, set by V ÷ R |
| “The resistance of the circuit went down” | Removing a branch raises the total resistance, from 6 Ω to 12 Ω |
| Correct: “p.d. across L1 is unchanged, so its current and brightness are unchanged” | Uses the quantity that is actually shared |
A real cell has a small internal resistance, which makes the terminal potential difference rise slightly when the load current falls. Unless a question mentions it, assume an ideal supply.
Check yourself
Three identical lamps are connected in parallel to a 12 V supply. Each draws 0.20 A. One lamp is removed. State the ammeter reading at the supply before and after, and what happens to the other two lamps.
Answer
Before: 3 × 0.20 = 0.60 A. After: 2 × 0.20 = 0.40 A.
The other two lamps still have 12 V across them, so each keeps a current of 0.20 A and the same brightness.
What to study next
Continue with comparing potential difference readings, then try the section practice set. Keep a note of each faulty explanation in the mistake log and paper-error review.
For a teacher to go through your explanations, see online one-to-one Physics tuition or try the one-hour trial class (from RM50).