Circuit solver (Kirchhoff's laws)
Describe a circuit part by part and get the current through and potential difference across every part, worked out with Kirchhoff’s laws - cells with internal resistance, resistors, capacitors, galvanometers and Wheatstone bridges - and the equivalent resistance between any two points.
Answer
- Equivalent between B and 0
- 2 Ω
Currents and potential differences
| Part | What | Between | Current (A) | p.d. (V) | Power (W) |
|---|---|---|---|---|---|
| E1 | cell | A – 0 | 2.4 | 9.6 | 28.8 |
| R1 | resistor | A – B | 2.4 | 4.8 | 11.52 |
| R2 | resistor | B – 0 | 0.8 | 4.8 | 3.84 |
| R3 | resistor | B – C | 1.6 | 4.8 | 7.68 |
| W1 | wire | C – 0 | 1.6 | 0 |
Worked out with Kirchhoff’s laws: the currents into each junction add up to zero, and so do the p.d.s round each loop.
The circuit
Potential at each point (point 0 at 0 V)
| Point | Potential (V) |
|---|---|
| 0 | 0 |
| A | 9.6 |
| B | 4.8 |
| C | 0 |
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