Battery internal resistance and a filament lamp load
A student investigates a battery using a variable resistor, an ammeter and a voltmeter connected so that V is the terminal voltage. At two settings the readings are (I = 0.10 A, V = 2.90 V) and (I = 0.50 A, V = 2.50 V). The resistor is then replaced by a small filament lamp whose characteristic the student had measured earlier: 0.60 V at 0.10 A, 1.30 V at 0.18 A, 2.00 V at 0.24 A, 2.70 V at 0.28 A and 3.00 V at 0.30 A. Using the dim end of this curve the student calculates R = 0.60/0.10 = 6.0 Ω and predicts a current of 3.0/(6.0 + 1.0) = 0.43 A. The ammeter actually reads 0.29 A. The student concludes that either the ammeter is faulty or that Ohm's law has failed. Assume the meters themselves draw negligible current.
1. Determine the emf and internal resistance of the battery from the two resistor readings, showing the reasoning.
Read the worked answer
Model the battery as an ideal emf E in series with internal resistance r. The terminal voltage is the emf minus the drop across r:
V = E − Ir
Substituting the two readings:
2.90 = E − 0.10r 2.50 = E − 0.50r
Subtracting the second from the first eliminates E:
0.40 = 0.40r → r = 1.0 Ω
Back-substituting: E = 2.90 + 0.10 × 1.0 = 3.00 V.
Note what the method relies on: only the change in V with I matters, so any constant voltmeter zero error would cancel in the subtraction. Two points are enough for a straight line, but a proper determination would plot V against I over many settings and use the intercept (E) and gradient (−r); with only two points there is no check that the V–I relation is actually linear, which matters at 0.50 A because heating of the electrolyte can raise r.
2. Find the actual operating point of the lamp in this circuit and explain why the prediction of 0.43 A fails.
Read the worked answer
The lamp and the battery must satisfy two conditions at once: the lamp's own characteristic, and the battery's load line V = 3.00 − 1.0I.
Compare point by point:
- I = 0.28 A: battery supplies 3.00 − 0.28 = 2.72 V; the lamp needs 2.70 V. The supply is slightly greater, so the current must be a little above 0.28 A.
- I = 0.30 A: battery supplies 3.00 − 0.30 = 2.70 V; the lamp needs 3.00 V. The supply is far too small, so the current must be below 0.30 A.
Intersection: I ≈ 0.29 A, V ≈ 2.71 V, which matches the measured 0.29 A.
The 0.43 A prediction fails because it treats the lamp as a fixed 6.0 Ω resistor. That 6.0 Ω belongs to the filament at 0.10 A, when it is barely warm. At 0.29 A the filament is far hotter, its resistivity is much larger, and the actual ratio V/I ≈ 2.71/0.29 ≈ 9.3 Ω. A component whose resistance depends on the current through it cannot be solved by substituting a single R into I = E/(R + r); the equation is not wrong, the assumed constant is.
3. The student claims the gap between 0.43 A and 0.29 A proves that Kirchhoff's laws do not hold for non-ohmic components. Evaluate this claim.
Read the worked answer
The claim is wrong, and the error is one of interpretation rather than of physics. Kirchhoff's laws are statements about conservation of charge (junction rule) and conservation of energy (loop rule). They hold for every circuit, ohmic or not - the loop rule is simply "the potential changes around a closed loop sum to zero".
What fails is a model, not a law: the assumption R = constant. Ohm's law (V = IR with constant R) is an empirical property of certain materials at constant temperature, not a circuit law. A filament lamp is a resistor but not an ohmic one; it still obeys the loop rule, which is exactly how the operating point was found (battery load line intersecting the lamp characteristic).
A useful test of the student's claim: the procedure that worked - solving the load line against the measured characteristic - is itself an application of the loop rule. If Kirchhoff's laws had failed, that method would not have reproduced the measured 0.29 A.
The genuine conclusion is narrower and more useful: for a non-ohmic component you must use its measured V–I characteristic (graphically or numerically), because the ratio V/I is a property of the particular operating point, not of the component.