D.C. Transients 253 For such a CR circuit the time constant, (Greek letter tau), is CRseconds. It may appear strange that the product of capacitance and resistance yields a result having units of time. This may be justiﬁ ed by considering a simple dimensional analysis, as follows. CQVItVRVICRItVVIt and so, secondsHence, secondsCR(8.2) Worked Example 8.1 Q An 8 μ F capacitor is connected in series with a 0.5 M resistor, across a 200 V d.c. supply. Calculate (a) the circuit time constant, (b) the initial charging current, (c) the p.d.s across the capacitor and resistor 4 seconds after the supply is connected. You may assume that the capacitor is initially fully discharged. A C 8 10 6 F; R 0.5 10 6 ; E 200 V 200 V0.5 MΩ8μF Fig. 8.8 (a) CR secondso s 800 5046611.Ans (b) II060200050400ER amptherefore A .1μ Ans (c) After seconds, voltV vEvvCCR0 6320 63220026 4...1AnsEEvvCR voltso V 20026 473 61..Ans 8.2 Capacitor-Resistor Series Circuit (Discharging) Consider the circuit of Fig. 8.1 , where the switch has been in position ‘ B ’ for sufﬁ cient time to allow the charging process to be completed.