Chapter 18 Kirchhoff Example RC Circuits

Find the currents in each branch of this circuit. I0I0 I1I1 I2I2 6  1  18 V 12  12 V + _ + _ 1  a f e c b d There are two junction points, so we can apply the junction rule ONCE. I 0 = I 1 + I 2

6  1  18 V 12  12 V + _ + _ 1  a f e c b d How many possible loops are there? b c d ae f c d a be f c b We have 3 unknowns, so we need 3 equations total. Therefore, we need use only 2 of the 3 equations provided by the loop rule. (The junction rule gave us 1 equation already!)

I0I0 6  1  18 V 12  12 V + _ + _ 1  a f e c b d I1I1 I2I2 Let’s go around bcda first. -(6  ) I V - (12  ) I 0 = 0 (12  ) I 0 + (6  ) I 1 = 18 V

I0I0 6  1  18 V 12  12 V + _ + _ 1  a f e c b d I1I1 I2I2 Let’s go around efcb second. 12 V - (1  ) I 2 + (6  ) I 1 - (1  ) I 2 = 0 (2  ) I 2 - (6  ) I 1 = 12 V

I0I0 6  1  18 V 12  12 V + _ + _ 1  a f e c b d I1I1 I2I2 I 0 = I 1 + I 2 (12  ) I 0 + (6  ) I 1 = 18 V (2  ) I 2 - (6  ) I 1 = 12 V I 0 = 2 A I 1 = - 1 A I 3 = 3 A (2  ) I 0 + (1  ) I 1 = 3 V(1  ) I 2 - (3  ) I 1 = 6 V

So far in Chapter 18, we’ve looked at circuits in which the currents are constant. These circuits have involved only resistors and sources of EMF. What happens if we put a capacitor into one of these circuits?

I V R + _ C If I start with an initially uncharged capacitor, what happens when I close the switch? Charge begins to build up on the capacitor. The more charge on the capacitor, the more work it takes to bring up the next bit of charge, the more slowly the capacitor will charge. When the capacitor is fully charged (i.e. Q = CV), the current will stop flowing.

q t Q=CV Q=0.632CV  V R + _ C  = the time constant = R C Q = final charge on C Charging a capacitor

R C,V 0,Q 0 Discharging a capacitor I start with a capacitor with charge Q 0 on it. What happens when I close the switch this time? Charge begins to flow off the capacitor through the resistor. Current is proportional to the potential drop across the capacitor. The less charge on the capacitor, the smaller the voltage drop, the smaller the current. So I expect the magnitude of the current to decrease as the capacitor discharges, eventually reaching 0.

Discharging a capacitor R C I q t Q 0 =CV 0 Q=0.368CV 0   = the time constant = R C Q 0 = initial charge on C

 = R C You might be saying to yourself, “I have a hard time believing that R times C gives me units of time!” [  ] = [R] [C] [  ] =  F = (V/A) (C/V) = C/A = C/(C/s) Yup! [t] = seconds!

6V 100  + _ 2f2f What is the initial current in this circuit after the switch is closed?

6V 100  + _ 2f2f What is the final charge on the capacitor? What is the time constant for this circuit?

6V 100  + _ 2f2f How long after the switch is closed does it take for the capacitor to become 90% charged? q = 0.9 Q f = 0.9 (12  C) = 10.8  C q (t) = Q f (1 - e -t/  ) 0.9 = (1 - e -t/200  s ) Take the natural log of both sides = e -t/200  s

6V 100  + _ 2f2f ln (0.1) = ln(e -t/200  s ) -2.3 = -t / 200  s t = 460  s