Consider the circuit shown in (Figure 1) . The batteries have emfs of E1=9.0V an
ID: 1319933 • Letter: C
Question
Consider the circuit shown in (Figure 1) . The batteries have emfs of E1=9.0V and E2=12.0V and the resistors have values of R1=28?, R2=63?, and R3=33?.
part a.
Determine the magnitudes of the currents in each resistor shown in the figure. Ignore internal resistance of the batteries.
I1, I2, I3=?
part b
Determine the magnitudes of the currents in each resistor shown in the figure. Assume that each battery has internal resistance
r = 1.0 ?.
I1, I2, I3 =?
note - solve both part a and b and be clear with your answers.
Explanation / Answer
(a)
Assuming that the negative terminals of both cells are connected together,
let a current of "a " pass clockwise in the dieagram through R1.
Let a current of "b" pass anticlockwise in the diagram through R2.
Both a and b pass through R2.
28a + (a + b)63 = 9
91a + 63b = 9-------------1
33b + (a + b)63 = 12
63a + 96b = 12
21a + 32b = 4-------------2
Multiplying 1 by 3 and 2 by 13
273a + 189b = 27-------------3
273a + 416b = 52-------------4
Apply (4) - (3)
227b = 25
b = 0.11 A
putting this value in (2)
21a + 32*0.11 = 4
a = 0.023 A
a + b = 0.11 + 0.023 = 0.133 A
a = 0.023 through R1
a + b = 0.133 through R2
b = 0.11 A through R3
(b)
Set the node shared by both batteries at 0V, arbitrarily. (You get to do that with exactly one node of your choice.)
Then the only unknown node voltage is the one shared by all three resistors. Call it Vx. Then:
Vx ? 28 + Vx ? 63 + Vx ? 33 = 9/28 + 0/63 + 12/33
Since we have internal resistance of 1 ohm for each battery here hence
The result goes like this:
Vx ? (28+1) + Vx ? (63+1) + Vx ? (33+1) = 9/(28+1) + 0/(63+1) + 12/(33+1)
Vx (1/29 + 1/64 + 1/34) = 9/29 + 12/34
Vx (0.0345 + 0.0156 + 0.0294) = 0.31 + 0.353
Vx * 0.0795 = 0.663
Vx = 8.34 V
So,
I(R1) = (9 - 8.34) ? (29) = 22.76 mA
I(R2) = (8.34 - 0) ? 63 = 130.3 mA
I(R3) = (12 - 8.34) ? (34) = 107.65 mA