Description
Problem 1. Consider the circuit presented in the figure below (E1 = 12 V, E2 = 8 V, r = 1 Ω, R = 8 Ω).
(a) Find the current through the resistor R,
(b) and the total rate of dissipation of electrical energy in the resistor R and in the internal
resistance of the batteries.
(c) In one of the batteries, chemical energy is being converted into electrical energy. In which
one it is happening, and at what rate?
(d) In one of the batteries, electrical energy is being converted into chemical energy. In which
one it is happening, and at what rate?
(e) Show that the overall rate of production of electrical energy is equal to the overall rate of
consumption of electrical energy in the circuit.
(5 × 1 marks)
Problem 2. For the circuit presented in the figure below, find the current through each of thee resistors. For
numerical calculations assume: R1 = 2 Ω, R2 = 4 Ω, R3 = 5 Ω, E1 = 20 V, E2 = 14 V, E3 = 36
V. The internal resistance of the emfs is negligible.
(4 marks)
Problem 3. For the circuit shown in the figure below what happens to the brightness of the bulbs when the
switch S is closed if the battery (a) has no internal resistance and (b) has non-negligible internal
resistance? Explain why.
(2 × 3/2 marks)
Problem 4. Strictly speaking, the formula q(t) = Qmaxe
−t/RC implies that an infinite amount of time is
required to discharge a capacitor in a R–C circuit completely. Yet for practical purposes, a
capacitor may be considered to be fully discharged after a finite time td, defined as the time
when the charge on the capacitor q(td) differs from zero by no more than the charge of one
electron.
(a) Find td if C = 0.92 µF, R = 670 kΩ, and Qmax = 7 µC.
(b) For a given Qmax is the time required to reach this state always the same number of time
constants, independent of R and C. Why or why not?
(1 + 2 marks)
Problem 5. Four resistors are connected to form a Wheatstone bridge – a circuit that can be used to measure
unknown resistance X, provided the resistances of N, M and P are known. The idea of the
measurement method is to tune (with the switches K1 and K2 closed) the variable resistance
X so that the potential difference between points b and c is zero and the galvanometer does
not show any current. The bridge is then said to be balanced. Show that in this configuration
X = MP/N.
(4 marks)
