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.
Currents and potential differences
| Part | What | Between | Current (A) | p.d. (V) | Power (W) |
|---|---|---|---|---|---|
| E1 | cell | P – 0 | 0.375 | 10 | 3.75 |
| R1 | resistor | P – A | 0.25 | 2.5 | 0.625 |
| R2 | resistor | P – B | 0.125 | 2.5 | 0.3125 |
| R3 | resistor | A – 0 | 0.25 | 7.5 | 1.875 |
| R4 | resistor | B – 0 | 0.125 | 7.5 | 0.9375 |
| G1 | galvanometer | A – B | 0 | 0 | 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.
Potential at each point (point 0 at 0 V)
| Point | Potential (V) |
|---|---|
| 0 | 0 |
| A | 7.5 |
| B | 7.5 |
| P | 10 |
Practise on real ISC questions
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