Exam Style Question - 4

QuestionA student connects a battery to a 4.0 Ω resistor and measures a terminal p.d. of 4.0 V. She replaces it with an 8.0 Ω resistor and the terminal p.d. rises to 4.8 V.(a) Calculate the current in each case.
(b) Determine the e.m.f. ε and the internal resistance r of the battery.
(c) Calculate the maximum power the battery can deliver to an external resistor.(a) Using I = V/R for the external resistor:I₁ = 4.0/4.0 = 1.0 A and I₂ = 4.8/8.0 = 0.60 A(b) Write ε = V + Ir for each case:ε = 4.0 + 1.0r
ε = 4.8 + 0.60rSubtracting the second equation from the first: 0 = −0.80 + 0.40r, so r = 2.0 Ω. Then ε = 4.0 + 1.0 × 2.0 = 6.0 V.Check: 6.0/(8.0 + 2.0) = 0.60 A. The terminal p.d. is higher with the larger resistor because the smaller current means fewer lost volts.(c) Maximum power is delivered when the external resistance equals the internal resistance, R = 2.0 Ω:I = ε/(R + r) = 6.0/4.0 = 1.5 A
P = I²R = 1.5² × 2.0 = 4.5 WThe same power is wasted inside the battery, so the efficiency at this point is only 50%.Exam tip: this is a graph in disguise. A plot of terminal p.d. V against current I is a straight line with gradient −r and y-intercept ε.