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asnl' 2'" 5'3” 175‘" J Irma—IL 26.34 Assume that the current ﬂows in the direction labelled on the graph. If we solve the equations and ﬁnd any current to be negative, it simply means that we assumed the
wrong direction for the current and we will correct it. Look at the junction where the voltage V is at. Use the junction rule: I2=I1+Is Via—V 52—1/ Ezi R1 “ R3 Vi127V 94/ Ei 35 n 25 V:%V%T.86V
751 I1 : (51 — V)/R1 : 0.046A
12 : V/Rg : 0.163A
13 = (52 i V)/R3 = 0.118A Our current directions as labelled in the graph happen to be the correct direction for
all three currents. (b) The internal resistances are in series with the batteries. If we look at the circuit, we
see that the ﬁrst battery’s internal resistor is in series with R1, the second battery’s is in series with R3. This means that with internal resistances, the equivalent circuit
is one that looks exactly like the circuit we have, with the batteries being perfect,
and with R1 being the original R1 plus r, and R3 similarly is the original R3 plus r. Therefore, we can simply use same equations as we did for part a, but with R1 =
25+1=26Q andR3=35+1=36Q I2:I1+Is
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13 : (52 — V)/R3 : 0.11m 26.64 I=V/R 5=IR+IT=V+%T Plug in the values for V and R: 9.7
5 Z W E?” 8.1
{S I W H?” Solve the system of equations, we get 5 = 112V and 7" = 5.39. 26.77 There are two possible answers because we don‘t know which direction the current is ﬂowing
in the circuit. First let‘s assume that the current given is ﬂowing to the right. The voltage drop across
the 4.0k resistor is V 2 IR 2 (3.10mA)(4.0kQ) = 12.4V, with the left side having higher
potential since current is ﬂowing towards the right. The current through the 8k resistor is thus I : V/R : DAV/81:9 : 1.55mA to the
right (since left side has higher potential). Which means that the current through the 5k
resistor is I : I1 + I2 : 3.10mA —— 1.55mA : 4.65mA to the right. So the voltage drop
across the 5k resistor is V 2 IR 2 (4.65mA)(5kQ) = 23.25V, with the left side being
higher potential. The voltage drop across the 19 resistor is V 2 IR 2 (4.65mA)(1Q) = 0.00465V. Since
the current ﬂowing through it is towards the left, its right side has higher potential than
the right side. Finally, putting everything together, we go around the circuit from a to b in counter—
clockwise loop, and we get Va + 0.00465V + 121/ + 1241/ + 23.25v : V5
V5,, : Vb — Va : 47.65V Now let‘s consider the case that the 3.10mA is ﬂowing through the 4k resistor towards
the left. The voltage drop across the 4k resistor (and the 8k, consequently) are still the
same 12.4V. But now, the right side has higher potential compared to the left side. The current through the 8k resistor is thus still 124/8000 = 1.55mA, but it’s towards
the left. The current through the 5k resistor is still I : I1 + I2 : 4.65mA, but it‘s also
towards the left. So voltage drop across the 5k resistor is still V : IR : (4.65mA)(5k§2) :
23.25V, but the right side has has potential now. Similarly, voltage drop across the 19 resistor is still V 2 IR 2 (4.65mA)(1Q) =
0.00465V, but as the current ﬂowing through it is now towards the right, the left side has
more potential now. So if we go around the same counterclockwise loop as before, from a to b, we get: Va 7 0.00465V + 12V 712.4V 7 23.25V = V}, In other words, except for the 12V battery, all other voltage drops have reversed their
direction because the current reversed direction. And the answer in this case is: vim : vb — va : 70.00465 + 12 — 12.4 — 23.25 : 723.65V ...
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 Spring '12
 welsch

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